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) b. ( 5 _ βˆ’1 w 3 ) _ Simplifying Exponential Expressions Recall that to simplify an expression means to rewrite it by combing terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions. Example 9 Simplifying Exponential Expressions Simplify each expression and write the answer with positive exponents only. a. (6m2nβˆ’1) 3 4 e. ( x2 √ β€” 2 ) ( x2 √ Solution b. 175 βˆ™ 17βˆ’4 βˆ™ 17βˆ’3 β€” βˆ’4 2 ) f. (3w2)5 _ (6wβˆ’2)2 c. ( 2 uβˆ’1v ) ____ v βˆ’1 d. (βˆ’2a3bβˆ’1)(5aβˆ’2b2) a. (6m2nβˆ’1) 3 = (6)3(m2)3(nβˆ’1)3 The power of a product rule = 63m2 βˆ™ 3nβˆ’1 βˆ™ 3 = 216m6nβˆ’3 216m6 _ n3 = b. 175 βˆ™ 17βˆ’4 βˆ™ 17βˆ’3 = 175 βˆ’ 4 βˆ’ 3 = 17βˆ’2 1 ___ = 172 or 1 ___ 289 c. ( uβˆ’1v _ vβˆ’1 ) 2 = = (uβˆ’1v)2 _ (vβˆ’1)2 uβˆ’2v2 _ vβˆ’2 = uβˆ’2v2βˆ’(βˆ’2) = uβˆ’2v 4 v 4 _ = u2 d. (βˆ’2a3bβˆ’1)(5aβˆ’2b2) = βˆ’2 βˆ™ 5 βˆ™ a3 βˆ™ aβˆ’2 βˆ™ bβˆ’1 βˆ™ b2 e10 βˆ™ a3 βˆ’ 2 βˆ™ bβˆ’1 + 2 = βˆ’10ab βˆ’. (3w2)5 _ (6wβˆ’2)2 = = 35 βˆ™ (w 2)5 _ 62 βˆ™ (wβˆ’2)2 35w 2 βˆ™ 5 _ 62 w βˆ’2 βˆ™ 2 243w10 _ 36w βˆ’#4 27w 10 βˆ’ (βˆ’4) _ 4 27w14 _ 4 = = = The power rule Simplify. The negative exponent rule The product rule Simplify. The negative exponent rule The power of a quotient rule The power of a product rule The quotient rule Simplify. The negative exponent rule Commut
ative and associative laws of multiplication The product rule Simplify. The product rule Simplify. The zero exponent rule The power of a product rule The power rule Simplify. The quotient rule and reduce fraction Simplify. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.2 EXPONENTS AND SCIENTIFIC NOTATION 25 Try It #9 Simplify each expression and write the answer with positive exponents only. a. (2uvβˆ’2)βˆ’3 b. x 8 Β· xβˆ’12 Β· x 2 c. ( e2f βˆ’3 _ f βˆ’1 ) d. (9r βˆ’5s3)(3r 6sβˆ’4) βˆ’3 4 twβˆ’2 ) e. ( _ 9 3 4 twβˆ’2 ) ( _ 9 f. (2h2k)4 _ (7hβˆ’1k2)2 Using Scientific Notation Recall at the beginning of the section that we found the number 1.3 Γ— 1013 when describing bits of information in digital images. Other extreme numbers include the width of a human hair, which is about 0.00005 m, and the radius of an electron, which is about 0.00000000000047 m. How can we effectively work read, compare, and calculate with numbers such as these? A shorthand method of writing very small and very large numbers is called scientific notation, in which we express numbers in terms of exponents of 10. To write a number in scientific notation, move the decimal point to the right of the first digit in the number. Write the digits as a decimal number between 1 and 10. Count the number of places n that you moved the decimal point. Multiply the decimal number by 10 raised to a power of n. If you moved the decimal left as in a very large number, n is positive. If you moved the decimal right as in a small large number, n is negative. For example, consider the number 2,780,418. Move the decimal left until it is to the right of the first nonzero digit, which is 2. 2,780418 6 places left 2.780418 We obtain 2.780418 by moving the decimal point 6 places to the left. Therefore, the exponent of 10 is 6, and it is positive because we moved the decimal point to the left. This is what we should expect for a large number. 2.780418
Γ— 106 Working with small numbers is similar. Take, for example, the radius of an electron, 0.00000000000047 m. Perform the same series of steps as above, except move the decimal point to the right. 0.00000000000047 00000000000004.7 13 places right Be careful not to include the leading 0 in your count. We move the decimal point 13 places to the right, so the exponent of 10 is 13. The exponent is negative because we moved the decimal point to the right. This is what we should expect for a small number. 4.7 Γ— 10βˆ’13 scientific notation A number is written in scientific notation if it is written in the form a Γ— 10n, where 1 ≀ |a| < 10 and n is an integer. Example 10 Converting Standard Notation to Scientific Notation Write each number in scientific notation. a. Distance to Andromeda Galaxy from Earth: 24,000,000,000,000,000,000,000 m b. Diameter of Andromeda Galaxy: 1,300,000,000,000,000,000,000 m c. Number of stars in Andromeda Galaxy: 1,000,000,000,000 d. Diameter of electron: 0.00000000000094 m e. Probability of being struck by lightning in any single year: 0.00000143 Solution a. 24,000,000,000,000,000,000,000 m ← 22 places 2.4 Γ— 1022 m b. 1,300,000,000,000,000,000,000 m ← 21 places 1.3 Γ— 1021 m Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 26 CHAPTER 1 PREREQUISITES c. 1,000,000,000,000 ← 12 places 1 Γ— 1012 d. 0.00000000000094 m β†’ 13 places 9.4 Γ— 10βˆ’13 m e. 0.00000143 β†’ 6 places 1.43 Γ— 10βˆ’6 Analysis Observe that, if the given number is greater than 1, as in examples a–c, the exponent of 10 is positive; and if the number is less than 1, as in examples d–e, the exponent is negative. Try It #10 Write each number in scientific notation. a. U.S. national debt per taxpayer (April 2014): $152,000 b. World population (April 2014
): 7,158,000,000 c. World gross national income (April 2014): $85,500,000,000,000 d. Time for light to travel 1 m: 0.00000000334 s e. Probability of winning lottery (match 6 of 49 possible numbers): 0.0000000715 Converting from Scientific to Standard Notation To convert a number in scientific notation to standard notation, simply reverse the process. Move the decimal n places to the right if n is positive or n places to the left if n is negative and add zeros as needed. Remember, if n is positive, the value of the number is greater than 1, and if n is negative, the value of the number is less than one. Example 11 Converting Scientific Notation to Standard Notation Convert each number in scientific notation to standard notation. a. 3.547 Γ— 1014 b. βˆ’2 Γ— 106 c. 7.91 Γ— 10βˆ’7 d. βˆ’8.05 Γ— 10βˆ’12 Solution a. 3.547 Γ— 1014 3.54700000000000 β†’ 14 places 354,700,000,000,000 b. βˆ’2 Γ— 106 βˆ’2.000000 β†’ 6 places βˆ’2,000,000 c. 7.91 Γ— 10βˆ’7 0000007.91 ← 7 places 0.000000791 d. βˆ’8.05 Γ— 10βˆ’12 βˆ’000000000008.05 ← 12 places βˆ’0.00000000000805 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.2 EXPONENTS AND SCIENTIFIC NOTATION 27 Try It #11 Convert each number in scientific notation to standard notation. a. 7.03 Γ— 105 c. βˆ’3.9 Γ— 10βˆ’13 b. βˆ’8.16 Γ— 1011 d. 8 Γ— 10βˆ’6 Using Scientific Notation in Applications Scientific notation, used with the rules of exponents, makes calculating with large or small numbers much easier than doing so using standard notation. For example, suppose we are asked to calculate the number of atoms in 1 L of water. Each water molecule contains 3 atoms (2 hydrogen and 1 oxygen). The average drop of water contains around 1.32 Γ— 1021 molecules of water and 1 L of water holds about 1.22 Γ— 104 average drops. Therefore, there are approximately 3 Γ— (1.32 Γ— 1021) Γ—
(1.22 Γ— 104) β‰ˆ 4.83 Γ— 1025 atoms in 1 L of water. We simply multiply the decimal terms and add the exponents. Imagine having to perform the calculation without using scientific notation! When performing calculations with scientific notation, be sure to write the answer in proper scientific notation. For example, consider the product (7 Γ— 104) Γ— (5 Γ— 106) = 35 Γ— 1010. The answer is not in proper scientific notation because 35 is greater than 10. Consider 35 as 3.5 Γ— 10. That adds a ten to the exponent of the answer. (35) Γ— 1010 = (3.5 Γ— 10) Γ— 1010 = 3.5 Γ— (10 Γ— 1010) = 3.5 Γ— 1011 Example 12 Using Scientific Notation Perform the operations and write the answer in scientific notation. a. (8.14 Γ— 10βˆ’7) (6.5 Γ— 1010) b. (4 Γ— 105) Γ· (βˆ’1.52 Γ— 109) c. (2.7 Γ— 105) (6.04 Γ— 1013) d. (1.2 Γ— 108) Γ· (9.6 Γ— 105) e. (3.33 Γ— 104) (βˆ’1.05 Γ— 107) (5.62 Γ— 105) Solution a. (8.14 Γ— 10βˆ’7) (6.5 Γ— 1010) = (8.14 Γ— 6.5) (10βˆ’7 Γ— 1010) = (52.91) (103) b. (4 Γ— 105) Γ· (βˆ’1.52 Γ— 109) = ( = 5.291 Γ— 104 4 ) ( ______ βˆ’1.52 105 ) ___ 109 β‰ˆ (βˆ’2.63) (10βˆ’4) = βˆ’2.63 Γ— 10βˆ’4 c. (2.7 Γ— 105) (6.04 Γ— 1013) = (2.7 Γ— 6.04) (105 Γ— 1013) = (16.308) (1018) d. (1.2 Γ— 108) Γ· (9.6 Γ— 105) = ( = 1.6308 Γ— 1019 108 ) ) ( ___ 105 = (0.125) (103) 1.2 ___ 9.6 = 1.25 Γ— 102 Commutative and associative properties of multiplication Product rule of exponents Scientific notation Commutative and associative properties of multiplication Quotient rule of exponents Scientific notation Commutative
and associative properties of multiplication Product rule of exponents Scientific notation Commutative and associative properties of multiplication Quotient rule of exponents Scientific notation e. (3.33 Γ— 104)(βˆ’1.05 Γ— 107) (5.62 Γ— 105) = [3.33 Γ— (βˆ’1.05) Γ— 5.62] (104 Γ— 107 Γ— 105) β‰ˆ (βˆ’19.65) (1016) = βˆ’1.965 Γ— 1017 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 28 CHAPTER 1 PREREQUISITES Try It #12 Perform the operations and write the answer in scientific notation. a. (βˆ’7.5 Γ— 108)(1.13 Γ— 10βˆ’2) b. (1.24 Γ— 1011) Γ· (1.55 Γ— 1018) c. (3.72 Γ— 109)(8 Γ— 103) d. (9.933 Γ— 1023) Γ· (βˆ’2.31 Γ— 1017) e. (βˆ’6.04 Γ— 109)(7.3 Γ— 102)(βˆ’2.81 Γ— 102) Example 13 Applying Scientific Notation to Solve Problems In April 2014, the population of the United States was about 308,000,000 people. The national debt was about $17,547,000,000,000. Write each number in scientific notation, rounding figures to two decimal places, and find the amount of the debt per U.S. citizen. Write the answer in both scientific and standard notations. Solution The population was 308,000,000 = 3.08 Γ— 108. The national debt was $17,547,000,000,000 β‰ˆ $1.75 Γ— 1013. To find the amount of debt per citizen, divide the national debt by the number of citizens. (1.75 Γ— 1013) Γ· (3.08 Γ— 108) = ( 1013 ) ____ 108 1.75 ____ 3.08 ) Γ— ( β‰ˆ 0.57 Γ— 105 The debt per citizen at the time was about $5.7 Γ— 104, or $57,000. = 5.7 Γ— 104 Try It #13 An average human body contains around 30,000,000,000,000 red blood cells. Each cell measures approximately 0.000008 m long. Write each number in scientific notation and find the total length if the cells were laid end
-to-end. Write the answer in both scientific and standard notations. Access these online resources for additional instruction and practice with exponents and scientific notation. β€’ Exponential Notation (http://openstaxcollege.org/l/exponnot) β€’ Properties of Exponents (http://openstaxcollege.org/l/exponprops) β€’ Zero Exponent (http://openstaxcollege.org/l/zeroexponent) β€’ Simplify Exponent Expressions (http://openstaxcollege.org/l/exponexpres) β€’ Quotient Rule for Exponents (http://openstaxcollege.org/l/quotofexpon) β€’ Scientific Notation (http://openstaxcollege.org/l/scientificnota) β€’ Converting to Decimal Notation (http://openstaxcollege.org/l/decimalnota) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.2 SECTION EXERCISES 29 1.2 SECTION EXERCISES VERBAL 1. Is 23 the same as 32? Explain. 3. What is the purpose of scientific notation? 2. When can you add two exponents? 4. Explain what a negative exponent does. NUMERIC For the following exercises, simplify the given expression. Write answers with positive exponents. 8. 44 Γ· 4 7. 32 Β· 33 6. 15βˆ’2 5. 92 βˆ’2 9. (22) 2 13. (80) 10. (5 βˆ’ 8)0 14. 5βˆ’2 Γ· 52 11. 113 Γ· 114 12. 65 Β· 6βˆ’7 For the following exercises, write each expression with a single base. Do not simplify further. Write answers with positive exponents. 15. 42 Β· 43 Γ· 4βˆ’4 18. 106 Γ· (1010) 17. (123 Β· 12) 16. βˆ’2 10 612 ___ 69 19. 7βˆ’6 Β· 7βˆ’3 20. (33 Γ· 34) 5 For the following exercises, express the decimal in scientific notation. 21. 0.0000314 22. 148,000,000 For the following exercises, convert each number in scientific notation to standard notation. 23. 1.6 Γ— 1010 24. 9.8 Γ— 10βˆ’9 ALGEBRAIC For the following exercises,
simplify the given expression. Write answers with positive exponents. xβˆ’3 _ 2 25. a3a2 _ a 26. mn2 _ mβˆ’2 βˆ’5 28. ( y 2 ) 27. ( b3c4) 29. ab 2 Γ· dβˆ’3 33. pβˆ’4q2 _ p 2qβˆ’3 37. 52m Γ· 50m 41. (bβˆ’3c) 3 βˆ’1 30. (w0x5) 34. (l Γ— w)2 β€” 2 x ) 38. ( 16 √ _ yβˆ’1 42. (x 2y13 Γ· y0) 2 31. m4 ___ n0 3 35. (y7) Γ· x 14 39. 23 _____ (3a)βˆ’2 βˆ’2 43. (9z3) y 32. yβˆ’4 (y2) 2 2 a __ ) 36. ( 23 40. (ma6) 2 1 _ m3a2 REAL-WORLD APPLICATIONS 44. To reach escape velocity, a rocket must travel at the rate of 2.2 Γ— 106 ft/min. Rewrite the rate in standard notation. 45. A dime is the thinnest coin in U.S. currency. A dime’s thickness measures 2.2 Γ— 106 m. Rewrite the number in standard notation. 46. The average distance between Earth and the Sun is 92,960,000 mi. Rewrite the distance using scientific notation. 47. A terabyte is made of approximately 1,099,500,000,000 bytes. Rewrite in scientific notation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 30 CHAPTER 1 PREREQUISITES 49. One picometer is approximately 3.397 Γ— 10βˆ’11 in. Rewrite this length using standard notation. 48. The Gross Domestic Product (GDP) for the United States in the first quarter of 2014 was $1.71496 Γ— 1013. Rewrite the GDP in standard notation. 50. The value of the services sector of the U.S. economy in the first quarter of 2012 was $10,633.6 billion. Rewrite this amount in scientific notation. TECHNOLOGY For the following exercises, use a graphing calculator to simplify. Round the answers to the nearest hundredth. 2 51. ( 123 m33 ) ______ 4βˆ’3 52. 173 Γ·
152x3 EXTENSIONS For the following exercises, simplify the given expression. Write answers with positive exponents. 32 ) 53. ( __ a3 βˆ’2 2 a4 ) ( __ 22 54. (62 βˆ’ 24) βˆ’5 2 x __ Γ· ( ) y 55. m2n3 _____ a2cβˆ’3 Β· aβˆ’7nβˆ’2 ______ m2c4 56. ( x 6y 3 _____ Β· x 3yβˆ’3 y βˆ’7 ___ x βˆ’3 ) 10 2 57. ( (ab2c)βˆ’3 ) _______ bβˆ’3 58. Avogadro’s constant is used to calculate the number of particles in a mole. A mole is a basic unit in chemistry to measure the amount of a substance. The constant is 6.0221413 Γ— 1023. Write Avogadro’s constant in standard notation. 59. Planck’s constant is an important unit of measure in quantum physics. It describes the relationship between energy and frequency. The constant is written as 6.62606957 Γ— 10βˆ’34. Write Planck’s constant in standard notation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.3 RADICALS AND RATIONAL EXPRESSIONS 31 LEARNING OBJECTIVES In this section, you will: β€’ Evaluate square roots. β€’ Use the product rule to simplify square roots. β€’ Use the quotient rule to simplify square roots. β€’ Add and subtract square roots. β€’ Rationalize denominators. β€’ Use rational roots. 1. 3 RADICALS AND RATIONAL EXPRESSIONS A hardware store sells 16-ft ladders and 24-ft ladders. A window is located 12 feet above the ground. A ladder needs to be purchased that will reach the window from a point on the ground 5 feet from the building. To find out the length of ladder needed, we can draw a right triangle as shown in Figure 1, and use the Pythagorean Theorem. 12 feet c 5 feet Figure 1 a2 + b2 = c2 52 + 122 = c2 169 = c2 Now, we need to find out the length that, when squared, is 169, to determine which ladder to choose. In other words, we need to find a square root. In this section, we will investigate methods of finding solutions to problems such as this one. Evaluating Square Roots When
the square root of a number is squared, the result is the original number. Since 42 = 16, the square root of 16 is 4. The square root function is the inverse of the squaring function just as subtraction is the inverse of addition. To undo squaring, we take the square root. In general terms, if a is a positive real number, then the square root of a is a number that, when multiplied by itself, gives a. The square root could be positive or negative because multiplying two negative numbers gives a positive number. The principal square root is the nonnegative number that when multiplied by itself equals a. The square root obtained using a calculator is the principal square root. The principal square root of a is written as √ the radicand, and the entire expression is called a radical expression. β€” a. The symbol is called a radical, the term under the symbol is called Radical Radicand β€” 25 √ Radical expression principal square root The principal square root of a is the nonnegative number that, when multiplied by itself, equals a. It is written as a radical expression, with a symbol called a radical over the term called the radicand: √ a. β€” Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 32 CHAPTER 1 PREREQUISITES β€” 25 = Β±5? Q & A… Does √ No. Although both 52 and (βˆ’5)2 are 25, the radical symbol implies only a nonnegative root, the principal square root. The principal square root of 25 is √ 25 = 5. β€” Example 1 Evaluating Square Roots Evaluate each expression. β€” 100 b. √ a. √ β€” β€” 16 √ Solution c. √ β€” 25 + 144 d. √ β€” 49 βˆ’ √ β€” 81 β€” a. √ b. √ c. √ d. √ β€” β€” β€” 100 = 10 because 102 = 100 √ 16 = √ 25 + 144 = √ 49 βˆ’ √ β€” β€” β€” β€” 4 = 2 because 42 = 16 and 22 = 4 169 = 13 because 132 = 169 81 = 7 βˆ’ 9 = βˆ’2 because 72 = 49 and 92 = 81 Q & A… For √ No. √ β€” 25 + 144, can we find the square roots before adding? 144 = 5 + 12 = 17. This is not equivalent to √ 25 + √ β€” β€” β€” 25 +
144 = 13. The order of operations requires us to add the terms in the radicand before finding the square root. Try It #1 a. √ β€” 225 b. √ β€” β€” 81 √ c. √ β€” 25 βˆ’ 9 d. √ β€” 36 + √ β€” 121 Using the Product Rule to Simplify Square Roots To simplify a square root, we rewrite it such that there are no perfect squares in the radicand. There are several properties of square roots that allow us to simplify complicated radical expressions. The first rule we will look at is the product rule for simplifying square roots, which allows us to separate the square root of a product of two numbers 5. We can also use into the product of two separate rational expressions. For instance, we can rewrite √ the product rule to express the product of multiple radical expressions as a single radical expression. 15 as √ 3 βˆ™ √ β€” β€” β€” the product rule for simplifying square roots If a and b are nonnegative, the square root of the product ab is equal to the product of the square roots of a and b. β€” ab = √ β€” a Β· √ β€” b √ How To… Given a square root radical expression, use the product rule to simplify it. 1. Factor any perfect squares from the radicand. 2. Write the radical expression as a product of radical expressions. 3. Simplify. Example 2 Using the Product Rule to Simplify Square Roots Simplify the radical expression. a. √ β€” 300 b. √ β€” 162a5b4 Solution a. √ β€” β€” β€” 100 βˆ™ 3 100 βˆ™ √ 3 3 β€” √ 10 √ Factor perfect square from radicand. Write radical expression as product of radical expressions. Simplify. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.3 RADICALS AND RATIONAL EXPRESSIONS 33 b. √ β€” β€” 81a4b4 βˆ™ 2a 81a4b4 βˆ™ √ 2a 2a √ 9a2b2 √ β€” β€” Factor perfect square from radicand. Write radical expression as product of radical expressions. Simplify. Try It #2 Simplify √ β€” 50x2y3z. How To… Given the product of multiple radical expressions, use the product rule to combine them into one radical expression. 1. Express the product
of multiple radical expressions as a single radical expression. 2. Simplify. Example 3 Using the Product Rule to Simplify the Product of Multiple Square Roots Simplify the radical expression. β€” 12 βˆ™ √ β€” 3 √ Solution β€” 12 βˆ™ 3 β€” 36 √ √ 6 Express the product as a single radical expression. Simplify. Try It #3 Simplify √ β€” 50x Β· √ β€” 2x assuming x > 0. Using the Quotient Rule to Simplify Square Roots Just as we can rewrite the square root of a product as a product of square roots, so too can we rewrite the square root of a quotient as a quotient of square roots, using the quotient rule for simplifying square roots. It can be helpful to separate the numerator and denominator of a fraction under a radical so that we can take their square roots separately. ___ We can rewrite √ 5 __ 2 as β€” 5 √ ____. 2 √ β€” the quotient rule for simplifying square roots The square root of the quotient a _ is equal to the quotient of the square roots of a and b, where b β‰  0. b a √ ____ b √ ___ √ a __ b = β€” β€” How To… Given a radical expression, use the quotient rule to simplify it. 1. Write the radical expression as the quotient of two radical expression. 2. Simplify the numerator and denominator. Example 4 Using the Quotient Rule to Simplify Square Roots Simplify the radical expression. Solution β€” β€” 5 √ _____ 36 √ 5 √ ____ 6 β€” ____ 5 ___ 36 √ Write as quotient of two radical expressions. Simplify denominator. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 34 CHAPTER 1 PREREQUISITES Try It #4 ____ Simplify √ 2x2 ____ 9y4. Example 5 Using the Quotient Rule to Simplify an Expression with Two Square Roots Simplify the radical expression. Solution _______ 234x11y √ ______ 26x7y β€” 9x4 √ 3x 2 β€” 234x11y √ _________ 26x7y √ β€” Combine numerator and denominator into one radical expression. Simplify fraction. Simplify square root. Try It #5 Simplify β€” 9a5 b 14 √
________. β€” 3a4b5 √ Adding and Subtracting Square Roots We can add or subtract radical expressions only when they have the same radicand and when they have the same β€” 2. However, it is often possible to radical type such as square roots. For example, the sum of √ β€” simplify radical expressions, and that may change the radicand. The radical expression √ 18 can be written with a 2 in the radicand, as 3 √ β€” 2 and 3 √ β€” 2 is 4 √ β€” 2, so √ 2 + 3 √ 2 = 4 √ 18 = √ 2 + √ β€” 2. β€” β€” β€” β€” How To… Given a radical expression requiring addition or subtraction of square roots, solve. 1. Simplify each radical expression. 2. Add or subtract expressions with equal radicands. Example 6 Adding Square Roots Add 5 √ β€” 12 + 2 √ β€” 3. Solution We can rewrite 5 √ the expression becomes 5(2) √ 12 as 5 √ β€” β€” β€” 4 Β· 3. According the product rule, this becomes 5 √ β€” 4 √ β€” 3. The square root of √ β€” 4 is 2, so 3, which is 10 √ 3. Now the terms have the same radicand so we can add. β€” 10 √ β€” 3 + 2 √ β€” 3 = 12 √ β€” 3 Try It #6 Add √ β€” 5 + 6 √ β€” 20. Example 7 Subtracting Square Roots Subtract 20 √ β€” 72a3b4c βˆ’ 14 √ β€” 8a3b4c. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.3 RADICALS AND RATIONAL EXPRESSIONS 35 Solution Rewrite each term so they have equal radicands. 20 √ β€” 72a3b4c = 20 √ β€” β€” β€” 4 √ 9 √ 2 √ = 20(3)(2) ∣ a ∣ b2 √ 2ac = 120 ∣ a ∣ b2 √ β€” β€” a √ 2ac β€” β€” a2 √ β€” 2 (b2) √ β€” c 14 √ β€” 8a3b4c = 14 √ β€” a2 √ β€” 2 (b2) √ β€” c β€” β€” 2 √
4 √ = 14(2) ∣ a ∣ b2 √ = 28 ∣ a ∣ b2 √ β€” 2ac β€” a √ 2ac β€” Now the terms have the same radicand so we can subtract. β€” β€” 120 ∣ a ∣ b2 √ 2ac βˆ’ 28 ∣ a ∣ b2 √ 2ac = 92 ∣ a ∣ b2 √ β€” 2ac Try It #7 Subtract 3 √ β€” 80x βˆ’ 4 √ β€” 45x. Rationalizing Denominators When an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. We can remove radicals from the denominators of fractions using a process called rationalizing the denominator. We know that multiplying by 1 does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by the form of 1 that will eliminate the radical. For a denominator containing a single term, multiply by the radical in the denominator over itself. In other words, if the denominator is b √ β€” c, multiply by β€” c √ ____. β€” c √ For a denominator containing the sum of a rational and an irrational term, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign of the radical portion of the denominator. If the denominator is a + b √ c, then the conjugate is a βˆ’ b √ β€” c. β€” How To… Given an expression with a single square root radical term in the denominator, rationalize the denominator. 1. Multiply the numerator and denominator by the radical in the denominator. 2. Simplify. Example 8 Rationalizing a Denominator Containing a Single Term Write in simplest form. β€” 3 2 √ ______ 10 3 √ β€” Solution The radical in the denominator is √ β€” 10. So multiply the fraction by. Then simplify. β€” 10 √ _____ 10 √ β€” β€” 3 2 √ ______ Β· 10 3 √ β€” β€” β€” 10 √ _____ 10 √ β€” 30 2 √ ______ 30 30 √ _____ 15 β€” Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 36 Try
It #8 Write β€” 3 12 √ ______ in simplest form. β€” 2 √ CHAPTER 1 PREREQUISITES How To… Given an expression with a radical term and a constant in the denominator, rationalize the denominator. 1. Find the conjugate of the denominator. 2. Multiply the numerator and denominator by the conjugate. 3. Use the distributive property. 4. Simplify. Example 9 Rationalizing a Denominator Containing Two Terms Write 4 _______ 1 + √ β€” in simplest form. 5 Solution Begin by finding the conjugate of the denominator by writing the denominator and changing the sign. So the conjugate β€” 5. Then multiply the fraction by β€” 1 βˆ’ √ _______ 1 βˆ’ √ 5. β€” 5 4 _______ βˆ™ 1 + √ β€” 5 1 βˆ’ √ _______ ________ βˆ’4 β€” 5 βˆ’ 1 √ Use the distributive property. Simplify. of 1 + √ β€” 5 is 1 βˆ’ √ Try It #9 Write 7 _______ 2 + √ β€” in simplest form. 3 Using Rational Roots Although square roots are the most common rational roots, we can also find cuberoots, 4th roots, 5th roots, and more. Just as the square root function is the inverse of the squaring function, these roots are the inverse of their respective power functions. These functions can be useful when we need to determine the number that, when raised to a certain power, gives a certain number. Understanding n th Roots Suppose we know that a3 = 8. We want to find what number raised to the 3rd power is equal to 8. Since 23 = 8, we say that 2 is the cube root of 8. The nth root of a is a number that, when raised to the nth power, gives a. For example, βˆ’3 is the 5th root of βˆ’243 because (βˆ’3)5 = βˆ’243. If a is a real number with at least one nth root, then the principal nth root of a is the number with the same sign as a that, when raised to the nth power, equals a. The principal nth root of a is written as expression, n is called the index of the radical. β€” n √ a, where n is a positive integer greater than or equal to 2. In the radical principal nth root √ If a is a real number with at least
one nth root, then the principal nth root of a, written as the same sign as a that, when raised to the nth power, equals a. The index of the radical is n. n β€” a, is the number with Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.3 RADICALS AND RATIONAL EXPRESSIONS 37 Example 10 Simplifying n th Roots Simplify each of the following: β€” √ βˆ’32 b. a, 024 Solution 5 a. √ β€” βˆ’32 = βˆ’2 because (βˆ’2)5 = βˆ’32 c. βˆ’ 3 √ β€” 8x6 ___ 125 d. 8 √ 4 β€” 4 3 βˆ’ √ β€” 48 b. First, express the product as a single radical expression. 4 √ β€” 4,096 = 8 because 84 = 4,096 c. β€” 3 8x6 βˆ’ √ ________ 125 √ β€” 3 Write as quotient of two radical expressions. βˆ’2x 2 _____ 5 Simplify. 4 √ d Simplify to get equal radicands. Add. Try It #10 Simplify. 3 a. √ β€” βˆ’216 β€” 4 √ b. 3 √ 80 _______ 5 β€” 4 3 c. 6 √ β€” 9,000 + 7 √ 3 β€” 576 Using Rational Exponents Radical expressions can also be written without using the radical symbol. We can use rational (fractional) exponents. The index must be a positive integer. If the index n is even, then a cannot be negative We can also have rational exponents with numerators other than 1. In these cases, the exponent must be a fraction in lowest terms. We raise the base to a power and take an nth root. The numerator tells us the power and the denominator tells us the root √ β€” am All of the properties of exponents that we learned for integer exponents also hold for rational exponents. rational exponents Rational exponents are another way to express principal nth roots. The general form for converting between a radical expression with a radical symbol and one with a rational exponent is β€” am How To… Given an expression with a rational exponent, write the expression as a radical. 1. Determine the power by looking at the numerator of the exponent. 2. Determine the root by looking at the denominator of the exponent. 3. Using the base
as the radicand, raise the radicand to the power and use the root as the index. Example 11 Writing Rational Exponents as Radicals 2 __ Write 343 as a radical. Simplify. 3 Solution The 2 tells us the power and the 3 tells us the root. 2 = ( _ 343 3 √ 3 β€” 343 ) 2 = √ 3 β€” 3432 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 38 CHAPTER 1 PREREQUISITES β€” 3 √ 343 = 7 because 7 3 = 343. Because the cube root is easy to find, it is easiest to find the cube root before We know that squaring for this problem. In general, it is easier to find the root first and then raise it to a power. 2 _ = ( 3 343 2 = 72 = 49 343 ) √ β€” 3 Try It #11 5 _ Write 9 as a radical. Simplify. 2 Example 12 Writing Radicals as Rational Exponents 7 Write using a rational exponent. 4 _ β€” a2 √ Solution 2. We get 4 ___ _ The power is 2 and the root is 7, so the rational exponent will be 7 4 _ β€” a2 √ βˆ’2 _. = 4a 7 7. Using properties of exponents, we get 2 _ a 7 Try It #12 Write x √ β€” (5y)9 using a rational exponent. Example 13 Simplifying Rational Exponents Simplify: 3 _ 1 _ ) ( 3x a. 5 ( 2x ) 5 4 Solution 1 _ 3 __ 5 4 x a. 30 x 3 _ + 1 _ 5 4 30x 19 _ 20 30x 1 _ 2 9 b. ( ) ___ 16 ____ √ 9 ___ 16 β€” 9 √ _____ 16 √ 3 _ 4 β€” b. ( 1 __ βˆ’ 2 16 ) ___ 9 Multiply the coefficient. Use properties of exponents. Simplify. Use definition of negative exponents. Rewrite as a radical. Use the quotient rule. Simplify. Try It #13 1 _ 6 _ ). ( 14x 5 3 Simplify 8x Access these online resources for additional instruction and practice with radicals and rational exponents. β€’ Radicals (http://openstaxcollege.org/l/introradical) β€’ Rational Exponents (http://openstaxcollege.org/l/rationexpon) β€’ Simplify Radicals
(http://openstaxcollege.org/l/simpradical) β€’ Rationalize Denominator (http://openstaxcollege.org/l/rationdenom) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.3 SECTION EXERCISES 39 1.3 SECTION EXERCISES VERBAL 1. What does it mean when a radical does not have 2. Where would radicals come in the order of an index? Is the expression equal to the radicand? Explain. operations? Explain why. 3. Every number will have two square roots. What is 4. Can a radical with a negative radicand have a real the principal square root? square root? Why or why not? NUMERIC For the following exercises, simplify each expression. 5. √ β€” 256 9. √ β€” 196 13. √ ____ 81 ___ 5 18 ______ β€” 162 √ 17. 21. √ β€” 150 ____ 4 _ 225 25. √ 29. 8 ________ 1 βˆ’ √ 17 β€” β€” 4 √ 33. 15 √ 125 _________ 5 β€” 4 6. √ β€” β€” √ 256 7. √ β€” 4(9 + 16) 10. √ β€” 1 14. √ β€” 800 18. √ β€” 192 11. √ β€” 98 15. √ β€” 169 + √ β€” 144 19. 14 √ β€” 6 βˆ’ 6 √ β€” 24 β€” 121 β€” 8. √ 289 βˆ’ √ ___ 27 12. √ _ 64 ___ 8 16. √ _ 50 20. 15 √ β€” 5 + 7 √ β€” 45 ____ 96 _ 100 ____ 405 _ 324 22. √ 26. √ 23. ( √ β€” 42 ) ( √ β€” 30 ) 24. 12 √ β€” 3 βˆ’ 4 √ β€” 75 ____ 360 _ 361 27. √ 28. 5 _______ 1 + √ 3 β€” 4 30. √ β€” 16 3 31. √ β€” 128 + 3 √ 3 β€” 2 β€” βˆ’32 ____ 243 32. 5 √ 3 34. 3 √ β€” βˆ’432 + √ 3 β€” 16 ALGEBRAIC For the following exercises, simplify each expression. 35. √ β€” 400x4 36. √ β€” 4y2 37. √ β€” 49p 1 _ 38. (144
p2q6) 2 289 β€” 5 _ 2 39. m √ ______ 225x3 _ 49x ____ 32 _ 14d 43. √ 47. √ 40. 9 √ β€” 3m2 + √ β€” 27 41. 3 √ β€” ab2 βˆ’ b √ β€” a 44. 3 √ β€” 44z + √ β€” 99z 45. √ β€” 50y8 3 _ 48. q √ 2 β€” 63p 49. β€” √ 8 ________ 1 βˆ’ √ 3x β€” 3 _ 2 51. w √ β€” 3 _ 32 βˆ’ w 2 √ β€” 50 52. √ β€” 108x4 + √ β€” 27x4 53. β€” 12x √ ________ 3 2 + 2 √ β€” 55. √ β€” 125n10 56. √ ____ 42q ____ 36q3 57. √ ______ 81m _ 361m2 42. β€” 2n 4 √ _______ 16n4 √ β€” 46. √ β€” 490bc2 50. √ ______ 20 _ 121d4 54. √ β€” 147k3 58. √ β€” 72c βˆ’ 2 √ β€” 2c Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 40 CHAPTER 1 PREREQUISITES 59. √ ______ 144 _ 324d2 3 60. √ β€” 24x6 + √ 3 β€” 81x6 61. 4 √ β€” 162x6 _____ 16x4 3 62. √ β€” 64y 3 63. √ β€” 128z3 βˆ’ √ 3 β€” βˆ’16z3 5 64. √ β€” 1,024c10 REAL-WORLD APPLICATIONS 65. A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So the length of the guy wire can be found by 90,000 + 160,000. What is the length evaluating √ of the guy wire? β€” EXTENSIONS For the following exercises, simplify each expression. 67. β€” β€” 8 βˆ’ √ 16 √
___________ βˆ’ 16 68. 4 _________ 1 _ 3 8 66. A car accelerates at a rate of 6 βˆ’ β€” 4 √ ____ m/s2 where t t √ β€” is the time in seconds after the car moves from rest. Simplify the expression. β€” 69. mn3 √ ________ Β· cβˆ’3 a2 √ β€” aβˆ’7nβˆ’2 _______ m2c4 √ β€” 70. a _______ a βˆ’ √ c β€” 71. β€” β€” 64y + 4 √ x √ __ 128y √ β€” y β€” β€” 72. ( 250x2 √ ________ 100b3 √ )( β€” b 7 √ _______ 125x √ β€” ) 3 β€” ______________ β€” 256 64 + √ √ __ 256 64 + √ √ β€” β€” 4 73. √ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.4 POLYNOMIALS 41 LEARNING OBJECTIVES Identify the degree and leading coefficient of polynomials. In this section, you will: β€’ β€’ Add and subtract polynomials. β€’ Multiply polynomials. β€’ Use FOIL to multiply binomials. β€’ Perform operations with polynomials of several variables. 1. 4 POLYNOMIALS Earl is building a doghouse, whose front is in the shape of a square topped with a triangle. There will be a rectangular door through which the dog can enter and exit the house. Earl wants to find the area of the front of the doghouse so that he can purchase the correct amount of paint. Using the measurements of the front of the house, shown in Figure 1, we can create an expression that combines several variable terms, allowing us to solve this problem and others like it. 1 foot x First find the area of the square in square feet. Then find the area of the triangle in square feet. 3 2 feet 2x Figure 1 A = s2 = (2x)2 = 4x2 1 _ A = bh 2 3 1 ) (2x Next find the area of the rectangular door in square feet. A = lw = x Β· 1 = x The area of the front of the doghouse can be found by adding the areas of the square and the triangle, and then 3 1 _ _
x βˆ’ x ft 2, or 4x 2 + subtracting the area of the rectangle. When we do this, we get 4x 2 + x ft 2. 2 2 In this section, we will examine expressions such as this one, which combine several variable terms. Identifying the Degree and Leading Coefficient of Polynomials The formula just found is an example of a polynomial, which is a sum of or difference of terms, each consisting of a variable raised to a nonnegative integer power. A number multiplied by a variable raised to an exponent, such as 384Ο€, is known as a coefficient. Coefficients can be positive, negative, or zero, and can be whole numbers, decimals, or fractions. Each product ai x i, such as 384Ο€w, is a term of a polynomial. If a term does not contain a variable, it is called a constant. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 42 CHAPTER 1 PREREQUISITES A polynomial containing only one term, such as 5x 4, is called a monomial. A polynomial containing two terms, such as 2x βˆ’ 9, is called a binomial. A polynomial containing three terms, such as βˆ’3x 2 + 8x βˆ’ 7, is called a trinomial. We can find the degree of a polynomial by identifying the highest power of the variable that occurs in the polynomial. The term with the highest degree is called the leading term because it is usually written first. The coefficient of the leading term is called the leading coefficient. When a polynomial is written so that the powers are descending, we say that it is in standard form. Leading coefficient Degree an x n +... + a2 x 2 + a1 x + a0 Leading term polynomials A polynomial is an expression that can be written in the form an x n +... + a2 x 2 + a1 x + a0 Each real number ai is called a coefficient. The number a0 that is not multiplied by a variable is called a constant. Each product ai x i is a term of a polynomial. The highest power of the variable that occurs in the polynomial is called the degree of a polynomial. The leading term is the term with the highest power, and its coefficient is called the leading
coefficient. How To… Given a polynomial expression, identify the degree and leading coefficient. 1. Find the highest power of x to determine the degree. 2. Identify the term containing the highest power of x to find the leading term. 3. Identify the coefficient of the leading term. Example 1 Identifying the Degree and Leading Coefficient of a Polynomial For the following polynomials, identify the degree, the leading term, and the leading coefficient. a. 3 + 2x2 βˆ’ 4x3 b. 5t5 βˆ’ 2t3 + 7t c. 6p βˆ’ p3 βˆ’ 2 Solution a. The highest power of x is 3, so the degree is 3. The leading term is the term containing that degree, βˆ’4x 3. The leading coefficient is the coefficient of that term, βˆ’4. b. The highest power of t is 5, so the degree is 5. The leading term is the term containing that degree, 5t5. The leading coefficient is the coefficient of that term, 5. c. The highest power of p is 3, so the degree is 3. The leading term is the term containing that degree, βˆ’p3, The leading coefficient is the coefficient of that term, βˆ’1. Try It #1 Identify the degree, leading term, and leading coefficient of the polynomial 4x 2 βˆ’ x 6 + 2x βˆ’ 6. Adding and Subtracting Polynomials We can add and subtract polynomials by combining like terms, which are terms that contain the same variables raised to the same exponents. For example, 5x 2 and βˆ’2x 2 are like terms, and can be added to get 3x 2, but 3x and 3x 2 are not like terms, and therefore cannot be added. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.4 POLYNOMIALS 43 How To… Given multiple polynomials, add or subtract them to simplify the expressions. 1. Combine like terms. 2. Simplify and write in standard form. Example 2 Adding Polynomials Find the sum. (12x 2 + 9x βˆ’ 21) + (4x 3 + 8x 2 βˆ’ 5x + 20) Solution 4x 3 + (12x 2 + 8x 2) + (9x βˆ’ 5x) + (βˆ’21 + 20) Combine like terms
. 4x 3 + 20x 2 + 4x βˆ’ 1 Simplify. Analysis We can check our answers to these types of problems using a graphing calculator. To check, graph the problem as given along with the simplified answer. The two graphs should be equivalent. Be sure to use the same window to compare the graphs. Using different windows can make the expressions seem equivalent when they are not. Try It #2 Find the sum. (2x 3 + 5x 2 βˆ’ x + 1) + (2x 2 βˆ’ 3x βˆ’ 4) Example 3 Subtracting Polynomials Find the difference. (7x 4 βˆ’ x 2 + 6x + 1) βˆ’ (5x 3 βˆ’ 2x 2 + 3x + 2) Solution 7x 4 βˆ’ 5x 3 + (βˆ’x 2 + 2x 2) + (6x βˆ’ 3x) + (1 βˆ’ 2) Combine like terms. 7x 4 βˆ’ 5x 3 + x 2 + 3x βˆ’ 1 Simplify. Analysis Note that finding the difference between two polynomials is the same as adding the opposite of the second polynomial to the first. Try It #3 Find the difference. (βˆ’7x 3 βˆ’ 7x 2 + 6x βˆ’ 2) βˆ’ (4x 3 βˆ’ 6x 2 βˆ’ x + 7) Multiplying Polynomials Multiplying polynomials is a bit more challenging than adding and subtracting polynomials. We must use the distributive property to multiply each term in the first polynomial by each term in the second polynomial. We then combine like terms. We can also use a shortcut called the FOIL method when multiplying binomials. Certain special products follow patterns that we can memorize and use instead of multiplying the polynomials by hand each time. We will look at a variety of ways to multiply polynomials. Multiplying Polynomials Using the Distributive Property To multiply a number by a polynomial, we use the distributive property. The number must be distributed to each term of the polynomial. We can distribute the 2 in 2(x + 7) to obtain the equivalent expression 2x + 14. When multiplying polynomials, the distributive property allows us to multiply each term of the first polynomial by each term of the second. We then add the products together and combine like terms to simplify. Download the OpenStax text for free at http://cnx.org
/content/col11759/latest. 44 CHAPTER 1 PREREQUISITES How To… Given the multiplication of two polynomials, use the distributive property to simplify the expression. 1. Multiply each term of the first polynomial by each term of the second. 2. Combine like terms. 3. Simplify. Example 4 Multiplying Polynomials Using the Distributive Property Find the product. (2x + 1) (3x 2 βˆ’ x + 4) Solution 2x (3x 2 βˆ’ x + 4) + 1(3x 2 βˆ’ x + 4) (6x 3 βˆ’ 2x 2 + 8x ) + (3x 2 βˆ’ x + 4) Use the distributive property. Multiply. 6x 3 + (βˆ’2x 2 + 3x 2) + (8x βˆ’ x) + 4 Combine like terms. 6x 3 + x 2 + 7x + 4 Simplify. Analysis We can use a table to keep track of our work, as shown in Table 1. Write one polynomial across the top and the other down the side. For each box in the table, multiply the term for that row by the term for that column. Then add all of the terms together, combine like terms, and simplify. 2x +1 βˆ’x 3x 2 6x 3 βˆ’2x 2 βˆ’x 3x 2 Table 1 +4 8x 4 Try It #4 Find the product. (3x + 2) (x3 βˆ’ 4x2 + 7) Using FOIL to Multiply Binomials A shortcut called FOIL is sometimes used to find the product of two binomials. It is called FOIL because we multiply the first terms, the outer terms, the inner terms, and then the last terms of each binomial. First terms Last terms Inner terms Outer terms The FOIL method arises out of the distributive property. We are simply multiplying each term of the first binomial by each term of the second binomial, and then combining like terms. How To… Given two binomials, use FOIL to simplify the expression. 1. Multiply the first terms of each binomial. 2. Multiply the outer terms of the binomials. 3. Multiply the inner terms of the binomials. 4. Multiply the last terms of each binomial. 5. Add the products. 6. Combine like terms and simplify. Download
the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.4 POLYNOMIALS 45 Example 5 Using FOIL to Multiply Binomials Use FOIL to find the product. (2x βˆ’ 10)(3x + 3) Solution Find the product of the first terms. 2x βˆ’ 18 3x + 3 2x Β· 3x = 6x2 Find the product of the outer terms. 2x βˆ’ 18 3x + 3 2x Β· 3 = 6x Find the product of the inner terms. 2x βˆ’ 18 3x + 3 βˆ’18 Β· 3x = βˆ’54x Find the product of the last terms. 2x βˆ’ 18 3x + 3 βˆ’18 Β· 3 = βˆ’54 6x2 + 6x βˆ’ 54x βˆ’ 54 Add the products. 6x2 + (6x βˆ’ 54x) βˆ’ 54 Combine like terms. 6x2 βˆ’ 48x βˆ’ 54 Simplify. Try It #5 Use FOIL to find the product. (x + 7)(3x βˆ’ 5) Perfect Square Trinomials Certain binomial products have special forms. When a binomial is squared, the result is called a perfect square trinomial. We can find the square by multiplying the binomial by itself. However, there is a special form that each of these perfect square trinomials takes, and memorizing the form makes squaring binomials much easier and faster. Let’s look at a few perfect square trinomials to familiarize ourselves with the form. (x + 5)2 = x2 + 10x + 25 (x βˆ’ 3)2 = x2 βˆ’ 6x + 9 (4x βˆ’ 1)2 = 4x2 βˆ’ 8x + 1 Notice that the first term of each trinomial is the square of the first term of the binomial and, similarly, the last term of each trinomial is the square of the last term of the binomial. The middle term is double the product of the two terms. Lastly, we see that the first sign of the trinomial is the same as the sign of the binomial. perfect square trinomials When a binomial is squared, the result is the first term squared added to double the product of both terms and the last term squared. (x + a)2 = (x + a)(x + a) = x2 + 2ax + a2
Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 46 CHAPTER 1 PREREQUISITES How To… Given a binomial, square it using the formula for perfect square trinomials. 1. Square the first term of the binomial. 2. Square the last term of the binomial. 3. For the middle term of the trinomial, double the product of the two terms. 4. Add and simplify. Example 6 Expanding Perfect Squares Expand (3x βˆ’ 8)2. Solution Begin by squaring the first term and the last term. For the middle term of the trinomial, double the product of the two terms. (3x)2 βˆ’ 2(3x)(8) + (βˆ’8)2 9x2 βˆ’ 48x + 64 Simplify. Try It #6 Expand (4x βˆ’ 1)2. Difference of Squares Another special product is called the difference of squares, which occurs when we multiply a binomial by another binomial with the same terms but the opposite sign. Let’s see what happens when we multiply (x + 1)(x βˆ’ 1) using the FOIL method. (x + 1)(x βˆ’ 1) = x2 βˆ’ x + x βˆ’ 1 = x2 βˆ’ 1 The middle term drops out, resulting in a difference of squares. Just as we did with the perfect squares, let’s look at a few examples. (x + 5)(x βˆ’ 5) = x2 βˆ’ 25 (x + 11)(x βˆ’ 11) = x2 βˆ’ 121 (2x + 3)(2x βˆ’ 3) = 4x2 βˆ’ 9 Because the sign changes in the second binomial, the outer and inner terms cancel each other out, and we are left only with the square of the first term minus the square of the last term. Q & A… Is there a special form for the sum of squares? No. The difference of squares occurs because the opposite signs of the binomials cause the middle terms to disappear. There are no two binomials that multiply to equal a sum of squares. difference of squares When a binomial is multiplied by a binomial with the same terms separated by the opposite sign, the result is the square of the first term minus the square of the last term. (a + b)(a βˆ’ b) = a2 βˆ’ b2 How To… Given a binomial multiplied by a
binomial with the same terms but the opposite sign, find the difference of squares. 1. Square the first term of the binomials. 2. Square the last term of the binomials. 3. Subtract the square of the last term from the square of the first term. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.4 POLYNOMIALS 47 Example 7 Multiplying Binomials Resulting in a Difference of Squares Multiply (9x + 4)(9x βˆ’ 4). Solution Square the first term to get (9x)2 = 81x2. Square the last term to get 42 = 16. Subtract the square of the last term from the square of the first term to find the product of 81x2 βˆ’ 16. Try It #7 Multiply (2x + 7)(2x βˆ’ 7). Performing Operations with Polynomials of Several Variables We have looked at polynomials containing only one variable. However, a polynomial can contain several variables. All of the same rules apply when working with polynomials containing several variables. Consider an example: (a + 2b)(4a βˆ’ b βˆ’ c) a(4a βˆ’ b βˆ’ c) + 2b(4a βˆ’ b βˆ’ c) Use the distributive property. 4a2 βˆ’ ab βˆ’ ac + 8ab βˆ’ 2b2 βˆ’ 2bc Multiply. 4a2 + (βˆ’ ab + 8ab) βˆ’ ac βˆ’ 2b2 βˆ’ 2bc Combine like terms. 4a2 + 7ab βˆ’ ac βˆ’ 2bc βˆ’ 2b2 Simplify. Example 8 Multiplying Polynomials Containing Several Variables Multiply (x + 4)(3x βˆ’ 2y + 5). Solution Follow the same steps that we used to multiply polynomials containing only one variable. x(3x βˆ’ 2y + 5) + 4(3x βˆ’ 2y + 5) Use the distributive property. 3x2 βˆ’ 2xy + 5x + 12x βˆ’ 8y + 20 Multiply. 3x2 βˆ’ 2xy + (5x + 12x) βˆ’ 8y + 20 Combine like terms. 3x2 βˆ’ 2xy + 17x βˆ’ 8y + 20 Simplify. Try It #8 Multiply (3x βˆ’ 1)(2x + 7y βˆ’ 9). Access
these online resources for additional instruction and practice with polynomials. β€’ Adding and Subtracting Polynomials (http://openstaxcollege.org/l/addsubpoly) β€’ Multiplying Polynomials (http://openstaxcollege.org/l/multiplpoly) β€’ Special Products of Polynomials (http://openstaxcollege.org/l/specialpolyprod) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 48 CHAPTER 1 PREREQUISITES 1.4 SECTION EXERCISES VERBAL 1. Evaluate the following statement: The degree of a 2. Many times, multiplying two binomials with two polynomial in standard form is the exponent of the leading term. Explain why the statement is true or false. variables results in a trinomial. This is not the case when there is a difference of two squares. Explain why the product in this case is also a binomial. 3. You can multiply polynomials with any number of terms and any number of variables using four basic steps over and over until you reach the expanded polynomial. What are the four steps? 4. State whether the following statement is true and explain why or why not: A trinomial is always a higher degree than a monomial. ALGEBRAIC For the following exercises, identify the degree of the polynomial. 5. 7x βˆ’ 2x2 + 13 9. x2 + 4x + 4 6. 14m3 + m2 βˆ’ 16m + 8 10. 6y4 βˆ’ y5 + 3y βˆ’ 4 For the following exercises, find the sum or difference. 11. (12x2 + 3x) βˆ’(8x2 βˆ’19) 13. (6w2 + 24w + 24) βˆ’ (3w βˆ’ 6w + 3) 15. (11b4 βˆ’ 6b3 + 18b2 βˆ’ 4b + 8) βˆ’ (3b3 + 6b2 + 3b) For the following exercises, find the product. 17. (4x + 2) (6x βˆ’ 4) 21. (9v βˆ’ 11) (11v βˆ’ 9) 18. (14c2 + 4c) (2c2 βˆ’ 3c) 22. (4t2 + 7t) (βˆ’3t2 + 4) For the following exercises, expand the
binomial. 24. (4x + 5)2 28. (2m βˆ’ 3)2 25. (3y βˆ’ 7)2 29. (3y βˆ’ 6)2 For the following exercises, multiply the binomials. 32. (9a βˆ’ 4)(9a + 4) 31. (4c + 1)(4c βˆ’ 1) 36. (14p + 7)(14p βˆ’ 7) 35. (4 + 4m)(4 βˆ’ 4m) 7. βˆ’625a8 + 16b4 8. 200p βˆ’ 30p2m + 40m3 12. (4z3 + 8z2 βˆ’ z) + (βˆ’2z2 + z + 6) 14. (7a3 + 6a2 βˆ’ 4a βˆ’ 13) + (βˆ’3a3 βˆ’ 4a2 + 6a + 17) 16. (49p2 βˆ’ 25) + (16p4 βˆ’ 32p2 + 16) 19. (6b2 βˆ’ 6) (4b2 βˆ’ 4) 23. (8n βˆ’ 4) (n2 + 9) 20. (3d βˆ’ 5)(2d + 9) 26. (12 βˆ’ 4x)2 30. (9b + 1)2 27. (4p + 9)2 33. (15n βˆ’ 6)(15n + 6) 37. (11q βˆ’ 10)(11q + 10) 34. (25b + 2)(25b βˆ’ 2) For the following exercises, find the sum or difference. 38. (2x2 + 2x + 1) (4x βˆ’ 1) 41. (y βˆ’ 2) (y2 βˆ’ 4y βˆ’ 9) 44. (4m βˆ’ 13) (2m2 βˆ’ 7m + 9) 47. (4t βˆ’ 5u)2 50. (b2 βˆ’ 1) (a2 + 2ab + b2) 39. (4t2 + t βˆ’ 7) (4t2 βˆ’ 1) 42. (6k βˆ’ 5) (6k2 + 5k βˆ’ 1) 45. (a + b) (a βˆ’ b) 48. (9m + 4n βˆ’ 1) (2m + 8) 51. (4r βˆ’ d)(6r + 7d) 40. (x βˆ’ 1) (x2 βˆ’ 2x + 1) 43. (3p2 + 2p βˆ’ 10) (p βˆ’ 1
) 46. (4x βˆ’ 6y) (6x βˆ’ 4y) 49. (4t βˆ’ x)(t βˆ’ x + 1) 52. (x + y) (x2 βˆ’ xy + y2) REAL-WORLD APPLICATIONS 53. A developer wants to purchase a plot of land to build a house. The area of the plot can be described by the following expression: (4x + 1)(8x βˆ’ 3) where x is measured in meters. Multiply the binomials to find the area of the plot in standard form. EXTENSIONS For the following exercises, perform the given operations. 55. (4t βˆ’ 7)2 (2t + 1) βˆ’ (4t2 + 2t + 11) 57. (a2 + 4ac + 4c2) (a2 βˆ’ 4c2) 54. A prospective buyer wants to know how much grain a specific silo can hold. The area of the floor of the silo is (2x + 9)2. The height of the silo is 10x + 10, where x is measured in feet. Expand the square and multiply by the height to find the expression that shows how much grain the silo can hold. 56. (3b + 6)(3b βˆ’ 6) (9b2 βˆ’ 36) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.5 FACTORING POLYNOMIALS 49 LEARNING OBJECTIVES In this section, you will: β€’ Factor the greatest common factor of a polynomial. β€’ Factor a trinomial. β€’ Factor by grouping. β€’ Factor a perfect square trinomial. β€’ Factor a difference of squares. β€’ Factor the sum and difference of cubes. β€’ Factor expressions using fractional or negative exponents. 1. 5 FACTORING POLYNOMIALS Imagine that we are trying to find the area of a lawn so that we can determine how much grass seed to purchase. The lawn is the green portion in Figure 1. 4 4 4 4 10x Figure 1 4 6x The area of the entire region can be found using the formula for the area of a rectangle. A = lw = 10x Β· 6x = 60x2 units 2 The areas of the portions that do not require grass seed need to be subtracted from the area of the entire region. The two square regions each have an
area of A = s2 = 42 = 16 units 2. The other rectangular region has one side of length 10x βˆ’ 8 and one side of length 4, giving an area of A = lw = 4(10x βˆ’ 8) = 40x βˆ’ 32 units 2. So the region that must be subtracted has an area of 2(16) + 40x βˆ’ 32 = 40x units 2. The area of the region that requires grass seed is found by subtracting 60x2 βˆ’ 40x units 2. This area can also be expressed in factored form as 20x(3x βˆ’ 2) units 2. We can confirm that this is an equivalent expression by multiplying. Many polynomial expressions can be written in simpler forms by factoring. In this section, we will look at a variety of methods that can be used to factor polynomial expressions. Factoring the Greatest Common Factor of a Polynomial When we study fractions, we learn that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, 4 is the GCF of 16 and 20 because it is the largest number that divides evenly into both 16 and 20 The GCF of polynomials works the same way: 4x is the GCF of 16x and 20x 2 because it is the largest polynomial that divides evenly into both 16x and 20x 2. When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 50 CHAPTER 1 PREREQUISITES greatest common factor The greatest common factor (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials. How To… Given a polynomial expression, factor out the greatest common factor. 1. Identify the GCF of the coefficients. 2. Identify the GCF of the variables. 3. Combine to find the GCF of the expression. 4. Determine what the GCF needs to be multiplied by to obtain each term in the expression. 5. Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by. Example 1 Factoring the Greatest Common Factor Factor 6x 3y
3 + 45x 2y 2 + 21xy. Solution First, find the GCF of the expression. The GCF of 6, 45, and 21 is 3. The GCF of x3, x 2, and x is x. (Note that the GCF of a set of expressions in the form xn will always be the exponent of lowest degree.) And the GCF of y3, y 2, and y is y. Combine these to find the GCF of the polynomial, 3xy. Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that 3xy(2x2y2) = 6x3y3, 3xy(15xy) = 45x2y2, and 3xy(7) = 21xy. Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by. (3xy)(2x 2y 2 + 15xy + 7) Analysis After factoring, we can check our work by multiplying. Use the distributive property to confirm that (3xy)(2x2y2 + 15xy + 7) = 6x3y3 + 45x2y2 + 21xy. Try It #1 Factor x(b2 βˆ’ a) + 6(b2 βˆ’ a) by pulling out the GCF. Factoring a Trinomial with Leading Coefficient 1 Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial x 2 + 5x + 6 has a GCF of 1, but it can be written as the product of the factors (x + 2) and (x + 3). Trinomials of the form x 2 + bx + c can be factored by finding two numbers with a product of c and a sum of b. The trinomial x 2 + 10x + 16, for example, can be factored using the numbers 2 and 8 because the product of those numbers is 16 and their sum is 10. The trinomial can be rewritten as the product of (x + 2) and (x + 8). factoring a trinomial with leading coefficient 1 A trinomial of the form x 2 + bx + c can be written in factored form as (x + p)(x + q)
where pq = c and p + q = b. Q & A… Can every trinomial be factored as a product of binomials? No. Some polynomials cannot be factored. These polynomials are said to be prime. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.5 FACTORING POLYNOMIALS 51 How To… Given a trinomial in the form x 2 + bx + c, factor it. 1. List factors of c. 2. Find p and q, a pair of factors of c with a sum of b. 3. Write the factored expression (x + p)(x + q). Example 2 Factoring a Trinomial with Leading Coefficient 1 Factor x 2 + 2x βˆ’ 15. Solution We have a trinomial with leading coefficient 1, b = 2, and c = βˆ’15. We need to find two numbers with a product of βˆ’15 and a sum of 2. In Table 1, we list factors until we find a pair with the desired sum. Factors of βˆ’15 1, βˆ’15 βˆ’1, 15 3, βˆ’5 βˆ’3, 5 Table 1 Sum of Factors βˆ’14 14 βˆ’2 2 Now that we have identified p and q as βˆ’3 and 5, write the factored form as (x βˆ’ 3)(x + 5). Analysis We can check our work by multiplying. Use FOIL to confirm that (x βˆ’ 3)(x + 5) = x2 + 2x βˆ’ 15. Q & A… Does the order of the factors matter? No. Multiplication is commutative, so the order of the factors does not matter. Try It #2 Factor x 2 βˆ’ 7x + 6. Factoring by Grouping Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial 2x 2 + 5x + 3 can be rewritten as (2x + 3)(x + 1) using this process. We begin by rewriting the original expression as 2x 2 + 2x + 3x + 3 and then factor each portion of the expression to obtain 2x(x + 1) +
3(x + 1). We then pull out the GCF of (x + 1) to find the factored expression. factor by grouping To factor a trinomial in the form ax 2 + bx + c by grouping, we find two numbers with a product of ac and a sum of b. We use these numbers to divide the x term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression. How To… Given a trinomial in the form ax 2 + bx + c, factor by grouping. 1. List factors of ac. 2. Find p and q, a pair of factors of ac with a sum of b. 3. Rewrite the original expression as ax 2 + px + qx + c. 4. Pull out the GCF of ax 2 + px. 5. Pull out the GCF of qx + c. 6. Factor out the GCF of the expression. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 52 CHAPTER 1 PREREQUISITES Example 3 Factoring a Trinomial by Grouping Factor 5x 2 + 7x βˆ’ 6 by grouping. Solution We have a trinomial with a = 5, b = 7, and c = βˆ’6. First, determine ac = βˆ’30. We need to find two numbers with a product of βˆ’30 and a sum of 7. In Table 2, we list factors until we find a pair with the desired sum. Factors of βˆ’30 1, βˆ’30 βˆ’1, 30 2, βˆ’15 βˆ’2, 15 3, βˆ’10 βˆ’3, 10 Table 2 Sum of Factors βˆ’29 29 βˆ’13 13 βˆ’7 7 So p = βˆ’3 and q = 10. 5x2 βˆ’ 3x + 10x βˆ’ 6 x(5x βˆ’ 3) + 2(5x βˆ’ 3) (5x βˆ’ 3)(x + 2) Rewrite the original expression as ax2 + px + qx + c. Factor out the GCF of each part. Factor out the GCF of the expression. Analysis We can check our work by multiplying. Use FOIL to confirm that (5x βˆ’ 3)(x + 2) = 5x2 + 7x βˆ’ 6. Try It #3 Factor. a. 2x 2 + 9x + 9 b. 6x2 + x βˆ’ 1 Factoring a
Perfect Square Trinomial A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term. a2 + 2ab + b2 = (a + b)2 and a2 βˆ’ 2ab + b2 = (a βˆ’ b)2 We can use this equation to factor any perfect square trinomial. perfect square trinomials A perfect square trinomial can be written as the square of a binomial: a2 + 2ab + b2 = (a + b)2 How To… Given a perfect square trinomial, factor it into the square of a binomial. 1. Confirm that the first and last term are perfect squares. 2. Confirm that the middle term is twice the product of ab. 3. Write the factored form as (a + b)2. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.5 FACTORING POLYNOMIALS 53 Example 4 Factoring a Perfect Square Trinomial Factor 25x 2 + 20x + 4. Solution Notice that 25x 2 and 4 are perfect squares because 25x 2 = (5x)2 and 4 = 22. Then check to see if the middle term is twice the product of 5x and 2. The middle term is, indeed, twice the product: 2(5x)(2) = 20x. Therefore, the trinomial is a perfect square trinomial and can be written as (5x + 2)2. Try It #4 Factor 49x 2 βˆ’ 14x + 1. Factoring a Difference of Squares A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied. We can use this equation to factor any differences of squares. a2 βˆ’ b2 = (a + b)(a βˆ’ b) differences of squares A difference of squares can be rewritten as two factors containing the same terms but opposite signs. a2 βˆ’ b2 = (a + b)(a βˆ’ b) How To… Given a difference of squares, factor it into binomials. 1. Confirm
that the first and last term are perfect squares. 2. Write the factored form as (a + b)(a βˆ’ b). Example 5 Factoring a Difference of Squares Factor 9x 2 βˆ’ 25. Solution Notice that 9x 2 and 25 are perfect squares because 9x 2 = (3x)2 and 25 = 52. The polynomial represents a difference of squares and can be rewritten as (3x + 5)(3x βˆ’ 5). Try It #5 Factor 81y2 βˆ’ 100. Q & A… Is there a formula to factor the sum of squares? No. A sum of squares cannot be factored. Factoring the Sum and Difference of Cubes Now, we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial. Similarly, the sum of cubes can be factored into a binomial and a trinomial, but with different signs. a3 βˆ’ b3 = (a βˆ’ b)(a2 + ab + b2) a3 + b3 = (a + b) (a2 βˆ’ ab + b2) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 54 CHAPTER 1 PREREQUISITES We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: Same Opposite Always Positive. For example, consider the following example. The sign of the first 2 is the same as the sign between x 3 βˆ’ 23. The sign of the 2x term is opposite the sign between x 3 βˆ’ 23. And the sign of the last term, 4, is always positive. x3 βˆ’ 23 = (x βˆ’ 2)(x2 + 2x + 4) sum and difference of cubes We can factor the sum of two cubes as We can factor the difference of two cubes as a3 + b3 = (a + b) (a2 βˆ’ ab + b2) a3 βˆ’ b3 = (a βˆ’ b)(a2 + ab + b2) How To… Given a sum of cubes or difference of cubes, factor it. 1. Confirm that the first and last term are cubes, a3 + b3 or a3 βˆ’ b3. 2. For a sum of cubes, write the
factored form as (a + b)(a2 βˆ’ ab + b2). For a difference of cubes, write the factored form as (a βˆ’ b)(a2 + ab + b2). Example 6 Factoring a Sum of Cubes Factor x3 + 512. Solution Notice that x 3 and 512 are cubes because 83 = 512. Rewrite the sum of cubes as (x + 8)(x2 βˆ’ 8x + 64). Analysis After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check. Try It #6 Factor the sum of cubes: 216a3 + b3. Example 7 Factoring a Difference of Cubes Factor 8x 3 βˆ’ 125. Solution Notice that 8x 3 and 125 are cubes because 8x 3 = (2x)3 and 125 = 53. Write the difference of cubes as (2x βˆ’ 5)(4x 2 + 10x + 25). Analysis Just as with the sum of cubes, we will not be able to further factor the trinomial portion. Try It #7 Factor the difference of cubes: 1,000x 3 βˆ’ 1. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.5 FACTORING POLYNOMIALS 55 Factoring Expressions with Fractional or Negative Exponents Expressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest 3 1 __ __ power. These expressions follow the same factoring rules as those with integer exponents. For instance, 2x can + __ __ __ and being rewritten as x be factored by pulling out x 2 4 4 Example 8 Factoring an Expression with Fractional or Negative Exponents 2 1 __ _ Factor 3x (x + 2) βˆ’ + 4(x + 2). 3 3 1 _ Solution Factor out the term with the lowest value of the exponent. In this case, that would be (x + 2) βˆ’. 3 1 _ (x + 2) βˆ’ (3x + 4(x + 2)) 3 1 _ (x + 2) βˆ’ (3x + 4x + 8) 3 1 _ (x
+ 2) βˆ’ (7x + 8) 3 Factor out the GCF. Simplify. Try It #8 3 1 __ _ + 7a (5a βˆ’ 1) βˆ’. Factor 2(5a βˆ’ 1) 4 4 Access these online resources for additional instruction and practice with factoring polynomials. β€’ Identify GCF (http://openstaxcollege.org/l/findgcftofact) β€’ Factor Trinomials when a Equals 1 (http://openstaxcollege.org/l/facttrinom1) β€’ Factor Trinomials when a is not equal to 1 (http://openstaxcollege.org/l/facttrinom2) β€’ Factor Sum or Difference of Cubes (http://openstaxcollege.org/l/sumdifcube) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 56 CHAPTER 1 PREREQUISITES 1.5 SECTION EXERCISES VERBAL 1. If the terms of a polynomial do not have a GCF, does 2. A polynomial is factorable, but it is not a perfect that mean it is not factorable? Explain. square trinomial or a difference of two squares. Can you factor the polynomial without finding the GCF? 3. How do you factor by grouping? ALGEBRAIC For the following exercises, find the greatest common factor. 4. 14x + 4xy βˆ’ 18xy2 6. 30x3y βˆ’ 45x2y2 + 135xy3 8. 36j4k2 βˆ’ 18j3k3 + 54j2k4 5. 49mb2 βˆ’ 35m2ba + 77ma2 7. 200p3m3 βˆ’ 30p2m3 + 40m3 9. 6y4 βˆ’ 2y3 + 3y2 βˆ’ y For the following exercises, factor by grouping. 11. 2a2 + 9a βˆ’ 18 10. 6x2 + 5x βˆ’ 4 14. 20w2 βˆ’ 47w + 24 15. 2p2 βˆ’ 5p βˆ’ 7 For the following exercises, factor the polynomial. 16. 7x2 + 48x βˆ’ 7 17. 10h2 βˆ’ 9h βˆ’ 9 12. 6c2 + 41c + 63 13. 6n2 βˆ’ 19n βˆ’ 11 18. 2b2 βˆ’
25b βˆ’ 247 19. 9d 2 βˆ’73d + 8 20. 90v2 βˆ’181v + 90 21. 12t2 + t βˆ’ 13 22. 2n2 βˆ’ n βˆ’ 15 24. 25y2 βˆ’ 196 25. 121p2 βˆ’ 169 26. 4m2 βˆ’ 9 23. 16x2 βˆ’ 100 27. 361d2 βˆ’ 81 28. 324x2 βˆ’ 121 29. 144b2 βˆ’ 25c2 30. 16a2 βˆ’ 8a + 1 31. 49n2 + 168n + 144 32. 121x2 βˆ’ 88x + 16 33. 225y2 + 120y + 16 34. m2 βˆ’ 20m + 100 35. m2 βˆ’ 20m + 100 36. 36q2 + 60q + 25 For the following exercises, factor the polynomials. 37. x3 + 216 38. 27y3 βˆ’ 8 39. 125a3 + 343 40. b3 βˆ’ 8d3 41. 64x3 βˆ’ 125 42. 729q3 + 1331 43. 125r3 + 1,728s3 1 2 __ _ 44. 4x(x βˆ’ 1) βˆ’ + 3(x βˆ’ 1) 3 3 4 1 __ __ + 7(10t + 3) 46. 3t(10t + 3) 3 3 1 6 __ __ βˆ’ 2(3y βˆ’ 13) 48. 9y(3y βˆ’ 13) 5 5 5 1 __ _ βˆ’ + 5(2d + 3) 50. 6d(2d + 3) 6 6 3 1 __ _ 45. 3c(2c + 3) βˆ’ βˆ’ 5(2c + 3) 4 4 3 2 __ _ 47. 14x(x + 2) βˆ’ + 5(x + 2) 5 5 3 1 _ 49. 5z(2z βˆ’ 9) βˆ’ _ + 11(2z βˆ’ 9) βˆ’ 2 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.5 SECTION EXERCISES 57 REAL-WORLD APPLICATIONS For the following exercises, consider this scenario: Charlotte has appointed a chairperson to lead a city beautification project. The first act is to install statues and fountains in one of the city’s parks. The park is a rectangle with an area of 98x 2 + 105x βˆ’ 27 m2, as shown in the following figure. The length
and width of the park are perfect factors of the area. l Γ— w = 98x 2 + 105x βˆ’ 27 51. Factor by grouping to find the length and width of the park. 53. At the northwest corner of the park, the city is going to install a fountain. The area of the base of the fountain is 9x2 βˆ’ 25 m2. Factor the area to find the lengths of the sides of the fountain. 52. A statue is to be placed in the center of the park. The area of the base of the statue is 4x 2 + 12x + 9 m2. Factor the area to find the lengths of the sides of the statue. For the following exercise, consider the following scenario: A school is installing a flagpole in the central plaza. The plaza is a square with side length 100 yd as shown in the figure below. The flagpole will take up a square plot with area x 2 βˆ’ 6x + 9 yd2. Area: x 2 βˆ’ 6x + 9 100 yards 100 yards 54. Find the length of the base of the flagpole by factoring. EXTENSIONS For the following exercises, factor the polynomials completely. 55. 16x 4 βˆ’ 200x2 + 625 56. 81y 4 βˆ’ 256 57. 16z4 βˆ’ 2,401a4 3 2 __ _ 58. 5x(3x + 2) βˆ’ + (12x + 8) 4 2 59. (32x 3 + 48x 2 βˆ’ 162x βˆ’ 243)βˆ’1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 58 CHAPTER 1 PREREQUISITES LEARNING OBJECTIVES In this section, you will: β€’ Simplify rational expressions. β€’ Multiply rational expressions. β€’ Divide rational expressions. β€’ Add and subtract rational expressions. β€’ Simplify complex rational expressions. 1. 6 RATIONAL EXPRESSIONS A pastry shop has fixed costs of $280 per week and variable costs of $9 per box of pastries. The shop’s costs per week in terms of x, the number of boxes made, is 280 + 9x. We can divide the costs per week by the number of boxes made to determine the cost per box of pastries. 280 + 9x _ x Notice that the result is a polynomial expression divided by a second polynomial expression. In this section, we will explore quotients of polyn
omial expressions. Simplifying Rational Expressions The quotient of two polynomial expressions is called a rational expression. We can apply the properties of fractions to rational expressions, such as simplifying the expressions by canceling common factors from the numerator and the denominator. To do this, we first need to factor both the numerator and denominator. Let’s start with the rational expression shown. x2 + 8x + 16 ____________ x2 + 11x + 28 We can factor the numerator and denominator to rewrite the expression. (x + 4)2 ____________ (x + 4)(x + 7) Then we can simplify that expression by canceling the common factor (x + 4). x + 4 _____ x + 7 How To… Given a rational expression, simplify it. 1. Factor the numerator and denominator. 2. Cancel any common factors. Example 1 Simplifying Rational Expressions Simplify x2 βˆ’ 9 __________ x2 + 4x + 3. Solution (x + 3)(x βˆ’ 3) ____________ (x + 3)(x + 1) x βˆ’ 3 _____ x + 1 Factor the numerator and the denominator. Cancel common factor (x + 3). Analysis We can cancel the common factor because any expression divided by itself is equal to 1. Q & A… Can the x2 term be cancelled in Example 1? No. A factor is an expression that is multiplied by another expression. The x 2 term is not a factor of the numerator or the denominator. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.6 RATIONAL EXPRESSIONS 59 Try It #1 Simplify x βˆ’ 6 _______. x 2 βˆ’ 36 Multiplying Rational Expressions Multiplication of rational expressions works the same way as multiplication of any other fractions. We multiply the numerators to find the numerator of the product, and then multiply the denominators to find the denominator of the product. Before multiplying, it is helpful to factor the numerators and denominators just as we did when simplifying rational expressions. We are often able to simplify the product of rational expressions. How To… Given two rational expressions, multiply them. 1. Factor the numerator and denominator. 2. Multiply the numerators. 3. Multiply the denominators. 4. Simplify. Example 2 Multiplying Rational Expressions Multip
ly the rational expressions and show the product in simplest form: x 2 + 4x βˆ’ 5 __ βˆ™ 3x + 18 2x βˆ’ 1 _______ x + 5 Solution (2x βˆ’ 1) (x + 5)(x βˆ’ 1) _______ ____________ βˆ™ (x + 5) 3(x + 6) (x + 5)(x βˆ’ 1)(2x βˆ’ 1) ___________________ 3(x + 6)(x + 5), (x + 5) (x βˆ’ 1)(2x βˆ’ 1) ___________________ 3(x + 6), (x + 5) (x βˆ’ 1)(2x βˆ’ 1) _____________ 3(x + 6) Factor the numerator and denominator. Multiply numerators and denominators. Cancel common factors to simplify. Try It #2 Multiply the rational expressions and show the product in simplest form: x 2 + 11x + 30 ____________ Β· x 2 + 5x + 6 x 2 + 7x + 12 ___________ x 2 + 8x + 16 Dividing Rational Expressions Division of rational expressions works the same way as division of other fractions. To divide a rational expression by another rational expression, multiply the first expression by the reciprocal of the second. Using this approach, we x2 Γ· x2. Once the division expression has been rewritten as a multiplication would rewrite as the product 3 expression, we can multiply as we did before. 3 1 __ __ Β· x2 x 3 __ = x3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 60 CHAPTER 1 PREREQUISITES How To… Given two rational expressions, divide them. 1. Rewrite as the first rational expression multiplied by the reciprocal of the second. 2. Factor the numerators and denominators. 3. Multiply the numerators. 4. Multiply the denominators. 5. Simplify. Example 3 Dividing Rational Expressions Divide the rational expressions and express the quotient in simplest form: 2x2 + x βˆ’ 6 __________ Γ· x2 βˆ’ 1 x2 βˆ’ 4 __________ x2 + 2x + 1 Solution 2x2 + x βˆ’#6 _____________ βˆ™ x2 βˆ’ 1 x2 + 2x + 1 __ x 2 βˆ’ 4 (x + 1)(x + 1) __ (x + 2)(x βˆ’ 2
) (2x βˆ’ 3)(x + 2) __ βˆ™ (x + 1)(x βˆ’ 1) (2x βˆ’ 3)(x + 2)(x + 1)(x + 1) ___ (x + 1)(x βˆ’ 1)(x + 2)(x βˆ’ 2) (2x βˆ’ 3)(x + 2)(x + 1)(x + 1) ___ (x + 1)(x βˆ’ 1)(x + 2)(x βˆ’ 2) (2x βˆ’ 3)(x + 1) __ (x βˆ’ 1)(x βˆ’ 2) Rewrite as multiplication. Factor the numerator and denominator. Multiply numerators and denominators. Cancel common factors to simplify. Try It #3 Divide the rational expressions and express the quotient in simplest form: 9x2 βˆ’ 16 __ 3x2 + 17x βˆ’ 28 Γ· 3x2 βˆ’ 2x βˆ’ 8 __ x2 + 5x βˆ’ 14 Adding and Subtracting Rational Expressions Adding and subtracting rational expressions works just like adding and subtracting numerical fractions. To add fractions, we need to find a common denominator. Let’s look at an example of fraction addition. 5 ___ 24 1 ___ 40 + = + 3 ___ 120 25 ___ 120 28 _ 120 = = 7 _ 30 We have to rewrite the fractions so they share a common denominator before we are able to add.We must do the same thing when adding or subtracting rational expressions. The easiest common denominator to use will be the least common denominator, or LCD. The LCD is the smallest multiple that the denominators have in common. To find the LCD of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, if the factored denominators were (x + 3)(x + 4) and (x + 4)(x + 5), then the LCD would be (x + 3)(x + 4)(x + 5). Once we find the LCD, we need to multiply each expression by the form of 1 that will change the denominator to the LCD. We would need to multiply the expression with a denominator of (x + 3)(x + 4) by x + 3 _ with a denominator of (x + 3)(x + 4) by. x + 3 and the expression x + 5 _ x + 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.6 RATIONAL
EXPRESSIONS 61 How To… Given two rational expressions, add or subtract them. 1. Factor the numerator and denominator. 2. Find the LCD of the expressions. 3. Multiply the expressions by a form of 1 that changes the denominators to the LCD. 4. Add or subtract the numerators. 5. Simplify. Example 4 Adding Rational Expressions Add the rational expressions: 6 5 _ y _ x + Solution First, we have to find the LCD. In this case, the LCD will be xy. We then multiply each expression by the appropriate form of 1 to obtain xy as the denominator for each fraction Β· 5y _ xy + 6x _ xy Now that the expressions have the same denominator, we simply add the numerators to find the sum. 6x + 5y _ xy y x _ x does not change the value of the original expression because any number divided by _ y or Analysis Multiplying by itself is 1, and multiplying an expression by 1 gives the original expression. Example 5 Subtracting Rational Expressions Subtract the rational expressions: Solution 6 __ βˆ’ x 2 + 4x + 4 2 _ x 2 βˆ’#4 6 _______ βˆ’ (x + 2)2 2 ____________ (x + 2)(x βˆ’ 2) Factor. 6 _______ (x + 2)2 βˆ™ x βˆ’ 2 _____ x βˆ’ 2 βˆ’ 2 ____________ βˆ™ (x + 2)(x βˆ’ 2) x + 2 _____ x + 2 6(x βˆ’ 2) _____________ (x + 2)2(x βˆ’ 2) βˆ’ 2(x + 2) _____________ (x + 2)2(x βˆ’ 2) 6x βˆ’ 12 βˆ’ (2x + 4) ________________ (x + 2)2(x βˆ’ 2) 4x βˆ’ 16 _____________ (x + 2)2(x βˆ’ 2) 4(x βˆ’ 4) _____________ (x + 2)2(x βˆ’ 2) Multiply each fraction to get LCD as denominator. Multiply. Apply distributive property. Subtract. Simplify. Q & A… Do we have to use the LCD to add or subtract rational expressions? No. Any common denominator will work, but it is easiest to use the LCD. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 62 CHAPTER 1
PREREQUISITES Try It #4 Subtract the rational expressions βˆ’#3 Simplifying Complex Rational Expressions A complex rational expression is a rational expression that contains additional rational expressions in the numerator, the denominator, or both. We can simplify complex rational expressions by rewriting the numerator and denominator a _ can be simplified by rewriting the as single rational expressions and dividing. The complex rational expression 1 _ + c b a _ numerator as the fraction and combining the expressions in the denominator as 1 1 + bc _. We can then rewrite the b a _ expression as a multiplication problem using the reciprocal of the denominator. We get βˆ™ 1, which is equal b _ 1 + bc to ab _. 1 + bc How To… Given a complex rational expression, simplify it. 1. Combine the expressions in the numerator into a single rational expression by adding or subtracting. 2. Combine the expressions in the denominator into a single rational expression by adding or subtracting. 3. Rewrite as the numerator divided by the denominator. 4. Rewrite as multiplication. 5. Multiply. 6. Simplify. Example 6 Simplifying Complex Rational Expressions Simplify Solution Begin by combining the expressions in the numerator into one expression. x 1 _ x _ x + y βˆ™ xy 1 _ x _ x + xy + 1 _ x x _ x to get LCD as denominator. Multiply by Add numerators. Now the numerator is a single rational expression and the denominator is a single rational expression. xy + 1 ______ x _______ x __ y We can rewrite this as division, and then multiplication. x _ y Γ· xy + 1 _ x xy + 1 y _ x _ βˆ™ x y(xy + 1) _ x 2 Rewrite as multiplication. Multiply. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.6 RATIONAL EXPRESSIONS 63 Try It #5 y x __ y βˆ’ __ x _______ y Simplify: Q & A… Can a complex rational expression always be simplified? Yes. We can always rewrite a complex rational expression as a simplified rational expression. Access these online resources for additional instruction and practice with rational expressions. β€’ Simplify Rational Expressions (http://openstaxcollege.org/l/simpratexpress) β€’ Multiply and Divide Rational Expressions (http
://openstaxcollege.org/l/multdivratex) β€’ Add and Subtract Rational Expressions (http://openstaxcollege.org/l/addsubratex) β€’ Simplify a Complex Fraction (http://openstaxcollege.org/l/complexfract) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 64 CHAPTER 1 PREREQUISITES 1.6 SECTION EXERCISES VERBAL 1. How can you use factoring to simplify rational 2. How do you use the LCD to combine two rational expressions? expressions? 3. Tell whether the following statement is true or false and explain why: You only need to find the LCD when adding or subtracting rational expressions. ALGEBRAIC For the following exercises, simplify the rational expressions. 4. x 2 βˆ’ 16 __________ x 2 βˆ’ 5x + 4 8. m βˆ’ 12 ________ m2 βˆ’ 144 5. y2 + 10y + 25 ____________ y2 + 11y + 30 6. 6a2 βˆ’ 24a + 24 _____________ 6a2 βˆ’ 24 7. 9b2 + 18b + 9 ____________ 3b + 3 9. 2x 2 + 7x βˆ’ 4 ___________ 4x 2 + 2x βˆ’ 2 10. 6x 2 + 5x βˆ’ 4 _____________ 3x 2 + 19x + 20 11. a2 + 9a + 18 ___________ a2 + 3a βˆ’ 18 12. 3c2 + 25c βˆ’ 18 ____________ 3c2 βˆ’ 23c + 14 13. 12n2 βˆ’ 29n βˆ’ 8 _____________ 28n2 βˆ’ 5n βˆ’ 3 For the following exercises, multiply the rational expressions and express the product in simplest form. 14. x 2 βˆ’ x βˆ’ 6 __________ Β· 2x 2 + x βˆ’ 6 2x 2 + 7x βˆ’ 15 ____________ x 2 βˆ’ 9 15. c2 + 2c βˆ’ 24 ___________ Β· c2 + 12c + 36 c2 βˆ’ 10c + 24 ____________ c2 βˆ’ 8c + 16 16. 2d 2 + 9d βˆ’ 35 ____________ Β· d 2 + 10d + 21 3d 2 + 2d βˆ’ 21 _____________ 3d 2 + 14d βˆ’ 49 17. 10h2 βˆ’ 9h βˆ’ 9 _____________ Β· 2h2 βˆ’
19h + 24 h2 βˆ’ 16h + 64 _____________ 5h2 βˆ’ 37h βˆ’ 24 18. 6b2 + 13b + 6 ____________ Β· 4b2 βˆ’ 9 6b2 + 31b βˆ’ 30 _____________ 18b2 βˆ’ 3b βˆ’ 10 19. 2d 2 + 15d + 25 _____________ Β· 4d 2 βˆ’ 25 2d 2 βˆ’ 15d + 25 _____________ 25d 2 βˆ’ 1 20. 6x 2 βˆ’ 5x βˆ’ 50 ______________ Β· 15x 2 βˆ’ 44x βˆ’ 20 20x 2 βˆ’ 7x βˆ’ 6 ____________ 2x 2 + 9x + 10 21. t 2 βˆ’ 1 _________ Β· t 2 + 4t + 3 t 2 + 2t βˆ’ 15 __________ t 2 βˆ’ 4t + 3 22. 2n2 βˆ’ n βˆ’ 15 ____________ Β· 6n2 + 13n βˆ’ 5 12n2 βˆ’ 13n + 3 _____________ 4n2 βˆ’ 15n + 9 For the following exercises, divide the rational expressions. 24. 3y2 βˆ’ 7y βˆ’ 6 ___________ Γ· 2y2 βˆ’ 3y βˆ’ 9 y2 + y βˆ’ 2 __________ 2y2 + y βˆ’ 3 26. q2 βˆ’ 9 __________ Γ· q2 + 6q + 9 q2 βˆ’ 2q βˆ’ 3 __________ q2 + 2q βˆ’ 3 28. 16x 2 + 18x βˆ’ 55 ______________ Γ· 32x 2 βˆ’ 36x βˆ’ 11 2x 2 + 17x + 30 _____________ 4x 2 + 25x + 6 30. 16a2 βˆ’ 24a + 9 _____________ Γ· 4a2 + 17a βˆ’ 15 16a2 βˆ’ 9 ____________ 4a2 + 11a + 6 32. 9x2 + 3x βˆ’ 20 ____________ Γ· 3x2 βˆ’ 7x + 4 6x2 + 4x βˆ’ 10 ____________ x2 βˆ’ 2x + 1 23. 36x 2 βˆ’ 25 _____________ Β· 6x 2 + 65x + 50 3x 2 + 32x + 20 ______________ 18x 2 + 27x + 10 25. 6p 2 + p βˆ’ 12 ____________ 8p 2 + 18p + 9 Γ· 6p 2 βˆ’ 11p + 4 ____________ 2p 2 + 11p βˆ’ 6 27. 18d 2 + 77d βˆ’ 18 ______________ Γ· 27
d 2 βˆ’ 15d + 2 3d 2 + 29d βˆ’ 44 _____________ 9d 2 βˆ’ 15d + 4 29. 144b2 βˆ’ 25 _____________ Γ· 72b2 βˆ’ 6b βˆ’ 10 18b2 βˆ’ 21b + 5 ______________ 36b2 βˆ’ 18b βˆ’ 10 31. 22y2 + 59y + 10 ______________ Γ· 12y2 + 28y βˆ’ 5 11y2 + 46y + 8 _____________ 24y2 βˆ’ 10y + 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.6 SECTION EXERCISES 65 For the following exercises, add and subtract the rational expressions, and then simplify. 4 __ + 33. x 10 ___ y 36. c + 2 _____ 3 βˆ’ c βˆ’ 4 _____ 4 34. βˆ’ 12 ___ 2q 6 ___ 3p 37. y + 3 _____ y βˆ’ 2 + y βˆ’ 3 _____ y + 1 35. 4 _____ a + 1 + 5 _____ a βˆ’ 3 38. x βˆ’ 1 _____ x + 1 βˆ’ 2x + 3 ______ 2x + 1 39. 3z _____ z + 1 + 2z + 5 ______ z βˆ’ 2 40. 4p _____ p + 1 βˆ’ p + 1 _____ 4p 41. x _____ x + 1 + y _____ y + 1 For the following exercises, simplify the rational expression. 42 45. 3 b _ a + _ 6 _______ 2b _ 3a 48. 4x _ 7 2x _ + 3 _ x _ 2 43 46 __ x βˆ’ 1 _ x + 1 49. c βˆ’ 1 _ c + 1 2c _ + c + 2 __ 2c + 1 _ c + 1 REAL-WORLD APPLICATIONS 51. Brenda is placing tile on her bathroom floor. The area of the floor is 15x2 βˆ’ 8x βˆ’ 7 ft 2. The area of one tile is x2 βˆ’ 2x + 1 ft 2. To find the number of tiles needed, simplify the rational expression: 15x2 βˆ’ 8x βˆ’ 7 __. x2 βˆ’ 2x + 1 44. p x _ _ βˆ’ 4 8 _______ p 47. 50 ab + Area = 15x2 βˆ’ 8x βˆ’ 7 52. The area of Sandy’s yard is 25x2 βˆ’ 625 ft 2. A patch of sod has an area of x2 βˆ’
10x + 25 ft 2. Divide the two areas and simplify to find how many pieces of sod Sandy needs to cover her yard. 53. Aaron wants to mulch his garden. His garden is x2 + 18x + 81 ft 2. One bag of mulch covers x2 βˆ’ 81 ft 2. Divide the expressions and simplify to find how many bags of mulch Aaron needs to mulch his garden. EXTENSIONS For the following exercises, perform the given operations and simplify. 54. x2 + x βˆ’ 6 __________ Β· x2 βˆ’ 2x βˆ’ 3 2x2 βˆ’ 3x βˆ’ 9 ___________ Γ· x2 βˆ’ x βˆ’ 2 10x2 + 27x + 18 ______________ x2 + 2x + 1 56. + 2a βˆ’ 3 ______ 2a + 3 4a + 1 ______ 2a βˆ’ 3 _______________ 4a2 + 9 _______ a 55. 2y2 βˆ’ 3y βˆ’ 20 3y2 βˆ’ 10y + 3 ____________ ____________ βˆ™ 2y2 βˆ’ y βˆ’ 15 3y2 + 5y βˆ’ 2 _________________________ y βˆ’ 4 57. x2 + 7x + 12 ___________ Γ· x2 + x βˆ’ 6 3x2 + 19x + 28 _____________ Γ· 8x2 βˆ’ 4x βˆ’ 24 2x2 + x βˆ’ 3 ___________ 3x2 + 4x βˆ’ 7 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 66 CHAPTER 1 PREREQUISITES CHAPTER 1 REVIEW Key Terms algebraic expression constants and variables combined using addition, subtraction, multiplication, and division associative property of addition the sum of three numbers may be grouped differently without affecting the result; in symbols, a + (b + c) = (a + b) + c associative property of multiplication the product of three numbers may be grouped differently without affecting the result; in symbols, a Β· (b Β· c) = (a Β· b) Β· c base in exponential notation, the expression that is being multiplied binomial a polynomial containing two terms coefficient any real number ai in a polynomial in the form an x n +... + a2 x 2 + a1 x + a0 commutative property of addition two numbers may be added in either order without affecting the result; in symbols, a + b = b + a commutative property of multiplication
two numbers may be multiplied in any order without affecting the result; in symbols, a Β· b = b Β· a constant a quantity that does not change value degree the highest power of the variable that occurs in a polynomial difference of squares the binomial that results when a binomial is multiplied by a binomial with the same terms, but the opposite sign distributive property the product of a factor times a sum is the sum of the factor times each term in the sum; in symbols, a Β· (b + c) = a Β· b + a Β· c equation a mathematical statement indicating that two expressions are equal exponent in exponential notation, the raised number or variable that indicates how many times the base is being multiplied exponential notation a shorthand method of writing products of the same factor factor by grouping a method for factoring a trinomial in the form ax 2 + bx + c by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression formula an equation expressing a relationship between constant and variable quantities greatest common factor the largest polynomial that divides evenly into each polynomial identity property of addition there is a unique number, called the additive identity, 0, which, when added to a number, results in the original number; in symbols, a + 0 = a identity property of multiplication there is a unique number, called the multiplicative identity, 1, which, when multiplied by a number, results in the original number; in symbols, a Β· 1 = a index the number above the radical sign indicating the nth root integers the set consisting of the natural numbers, their opposites, and 0: { …, βˆ’3, βˆ’2, βˆ’1, 0, 1, 2, 3,…} inverse property of addition for every real number a, there is a unique number, called the additive inverse (or opposite), denoted βˆ’a, which, when added to the original number, results in the additive identity, 0; in symbols, a + (βˆ’a) = 0 inverse property of multiplication for every non-zero real number a, there is a unique number, called the multiplicative 1 __ a, which, when multiplied by the original number, results in the multiplicative inverse (or reciprocal), denoted 1 __ a = 1 identity, 1; in symbols, a Β· irrational numbers the set of all numbers that are not rational; they cannot be written as either a terminating or repeating decimal; they cannot
be expressed as a fraction of two integers leading coefficient the coefficient of the leading term leading term the term containing the highest degree least common denominator the smallest multiple that two denominators have in common Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 1 REVIEW 67 monomial a polynomial containing one term natural numbers the set of counting numbers: {1, 2, 3,…} order of operations a set of rules governing how mathematical expressions are to be evaluated, assigning priorities to operations perfect square trinomial the trinomial that results when a binomial is squared polynomial a sum of terms each consisting of a variable raised to a nonnegative integer power principal nth root the number with the same sign as a that when raised to the nth power equals a principal square root the nonnegative square root of a number a that, when multiplied by itself, equals a radical the symbol used to indicate a root radical expression an expression containing a radical symbol radicand the number under the radical symbol rational expression the quotient of two polynomial expressions rational numbers the set of all numbers of the form m __ n, where m and n are integers and n β‰  0. Any rational number may be written as a fraction or a terminating or repeating decimal. real number line a horizontal line used to represent the real numbers. An arbitrary fixed point is chosen to represent 0; positive numbers lie to the right of 0 and negative numbers to the left. real numbers the sets of rational numbers and irrational numbers taken together scientific notation a shorthand notation for writing very large or very small numbers in the form a Γ— 10n where 1 ≀ ∣ a ∣ < 10 and n is an integer term of a polynomial any ai x i of a polynomial in the form anx n +... + a2 x 2 + a1 x + a0 trinomial a polynomial containing three terms variable a quantity that may change value whole numbers the set consisting of 0 plus the natural numbers: {0, 1, 2, 3,…} Key Equations Rules of Exponents For nonzero real numbers a and b and integers m and n Product rule Quotient rule Power rule Zero exponent rule Negative rule Power of a product rule Power of a quotient rule am βˆ™ an = am + n am ___ an = am βˆ’ n (am)n = am βˆ™ n a0 = 1 aβˆ’n = 1 __ an (
a βˆ™ b)n = an βˆ™ bn a __ ) ( b n = an __ bn perfect square trinomial (x + a)2 = (x + a)(x + a) = x2 + 2ax + a2 difference of squares difference of squares (a + b)(a βˆ’ b) = a2 βˆ’ b2 a2 βˆ’ b2 = (a + b)(a βˆ’ b) perfect square trinomial a2 + 2ab + b2 = (a + b)2 sum of cubes difference of cubes a3 + b3 = (a + b)(a2 βˆ’ ab + b2) a3 βˆ’ b3 = (a βˆ’ b)(a2 + ab + b2) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 68 CHAPTER 1 PREREQUISITES Key Concepts 1.1 Real Numbers: Algebra Essentials β€’ Rational numbers may be written as fractions or terminating or repeating decimals. See Example 1 and Example 2. β€’ Determine whether a number is rational or irrational by writing it as a decimal. See Example 3. β€’ The rational numbers and irrational numbers make up the set of real numbers. See Example 4. A number can be classified as natural, whole, integer, rational, or irrational. See Example 5. β€’ The order of operations is used to evaluate expressions. See Example 6. β€’ The real numbers under the operations of addition and multiplication obey basic rules, known as the properties of real numbers. These are the commutative properties, the associative properties, the distributive property, the identity properties, and the inverse properties. See Example 7. β€’ Algebraic expressions are composed of constants and variables that are combined using addition, subtraction, multiplication, and division. See Example 8. They take on a numerical value when evaluated by replacing variables with constants. See Example 9, Example 10, and Example 12. β€’ Formulas are equations in which one quantity is represented in terms of other quantities. They may be simplified or evaluated as any mathematical expression. See Example 11 and Example 13. 1.2 Exponents and Scientific Notation β€’ Products of exponential expressions with the same base can be simplified by adding exponents. See Example 1. β€’ Quotients of exponential expressions with the same base can be simplified by subtracting exponents. See Example 2. β€’ Powers of exponential expressions with the same base can be
simplified by multiplying exponents. See Example 3. β€’ An expression with exponent zero is defined as 1. See Example 4. β€’ An expression with a negative exponent is defined as a reciprocal. See Example 5 and Example 6. β€’ The power of a product of factors is the same as the product of the powers of the same factors. See Example 7. β€’ The power of a quotient of factors is the same as the quotient of the powers of the same factors. See Example 8. β€’ The rules for exponential expressions can be combined to simplify more complicated expressions. See Example 9. β€’ Scientific notation uses powers of 10 to simplify very large or very small numbers. See Example 10 and Example 11. β€’ Scientific notation may be used to simplify calculations with very large or very small numbers. See Example 12 and Example 13. 1.3 Radicals and Rational Expressions β€’ The principal square root of a number a is the nonnegative number that when multiplied by itself equals a. See Example 1. β€’ If a and b are nonnegative, the square root of the product ab is equal to the product of the square roots of a and b See Example 2 and Example 3. a __ is equal to the quotient of the square roots of a and b β€’ If a and b are nonnegative, the square root of the quotient b See Example 4 and Example 5. β€’ We can add and subtract radical expressions if they have the same radicand and the same index. See Example 6 and Example 7. β€’ Radical expressions written in simplest form do not contain a radical in the denominator. To eliminate the square root radical from the denominator, multiply both the numerator and the denominator by the conjugate of the denominator. See Example 8 and Example 9. β€’ The principal nth root of a is the number with the same sign as a that when raised to the nth power equals a. These roots have the same properties as square roots. See Example 10. β€’ Radicals can be rewritten as rational exponents and rational exponents can be rewritten as radicals. See Example 11 and Example 12. β€’ The properties of exponents apply to rational exponents. See Example 13. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 1 REVIEW 69 1.4 Polynomials β€’ A polynomial is a sum of terms each consisting of a variable raised to a non-negative integer power. The degree is the highest power of
the variable that occurs in the polynomial. The leading term is the term containing the highest degree, and the leading coefficient is the coefficient of that term. See Example 1. β€’ We can add and subtract polynomials by combining like terms. See Example 2 and Example 3. β€’ To multiply polynomials, use the distributive property to multiply each term in the first polynomial by each term in the second. Then add the products. See Example 4. β€’ FOIL (First, Outer, Inner, Last) is a shortcut that can be used to multiply binomials. See Example 5. β€’ Perfect square trinomials and difference of squares are special products. See Example 6 and Example 7. β€’ Follow the same rules to work with polynomials containing several variables. See Example 8. 1.5 Factoring Polynomials β€’ The greatest common factor, or GCF, can be factored out of a polynomial. Checking for a GCF should be the first step in any factoring problem. See Example 1. β€’ Trinomials with leading coefficient 1 can be factored by finding numbers that have a product of the third term and a sum of the second term. See Example 2. β€’ Trinomials can be factored using a process called factoring by grouping. See Example 3. β€’ Perfect square trinomials and the difference of squares are special products and can be factored using equations. See Example 4 and Example 5. β€’ The sum of cubes and the difference of cubes can be factored using equations. See Example 6 and Example 7. β€’ Polynomials containing fractional and negative exponents can be factored by pulling out a GCF. See Example 8. 1.6 Rational Expressions β€’ Rational expressions can be simplified by cancelling common factors in the numerator and denominator. See Example 1. β€’ We can multiply rational expressions by multiplying the numerators and multiplying the denominators. See Example 2. β€’ To divide rational expressions, multiply by the reciprocal of the second expression. See Example 3. β€’ Adding or subtracting rational expressions requires finding a common denominator. See Example 4 and Example 5. β€’ Complex rational expressions have fractions in the numerator or the denominator. These expressions can be simplified. See Example 6. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 7 0 CHAPTER 1 PREREQUISITES CHAPTER 1 REVIEW EXERCIS
ES REAL NUMBERS: ALGEBRA ESSENTIALS For the following exercises, perform the given operations. 1. (5 βˆ’ 3 Β· 2)2 βˆ’ 6 2. 64 Γ· (2 Β· 8) + 14 Γ· 7 3. 2 Β· 52 + 6 Γ· 2 For the following exercises, solve the equation. 4. 5x + 9 = βˆ’11 5. 2y + 42 = 64 For the following exercises, simplify the expression. 6. 9(y + 2) Γ· 3 Β· 2 + 1 7. 3m(4 + 7) βˆ’ m For the following exercises, identify the number as rational, irrational, whole, or natural. Choose the most descriptive answer. 8. 11 9. 0 5 __ 10. 6 11. √ β€” 11 EXPONENTS AND SCIENTIFIC NOTATION For the following exercises, simplify the expression. 12. 22 Β· 24 16. (xy)4 _ y3 Β· 2 _ x5 13. 45 __ 43 17. 4βˆ’2x 3yβˆ’3 _______ 2x 0 4 a2 b3 ) 14. ( __ βˆ’2 18. ( 2x2 _ y ) 15. 6a2 Β· a0 _ 2aβˆ’4 19. ( 16a3 b2 ) (4abβˆ’1)βˆ’2 ____ 20. Write the number in standard notation: 21. Write the number in scientific notation: 16,340,000 2.1314 Γ— 10βˆ’6 RADICALS AND RATIONAL EXPRESSIONS For the following exercises, find the principal square root. 22. √ β€” 121 26. √ β€” 162 23. √ β€” 196 27. √ ___ 32 _ 25 24. √ β€” 361 28. √ ___ 80 _ 81 25. √ β€” 75 29. √ _____ 49 _ 1250 30. 2 _______ 4 + √ 2 β€” 31 32. 12 √ β€” 5 βˆ’ 13 √ β€” 5 5 33. √ β€” βˆ’243 34. β€” 3 250 √ _ βˆ’8 √ β€” 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 1 REVIEW 71 POLYNOMIALS For the following exercises, perform the given operations and simplify. 35. (3x3 + 2x βˆ’ 1) + (4x 2 βˆ’ 2x + 7) 36. (2y + 1) βˆ’ (2y 2
βˆ’ 2y βˆ’ 5) 37. (2x 2 + 3x βˆ’ 6) + (3x 2 βˆ’ 4x + 9) 38. (6a2 + 3a + 10) βˆ’ (6a2 βˆ’3a + 5) 39. (k + 3)(k βˆ’ 6) 41. (x + 1)(x 2 + 1) 43. (a + 2b)(3a βˆ’ b) 40. (2h + 1)(3h βˆ’ 2) 42. (m βˆ’ 2)(m2 + 2m βˆ’ 3) 44. (x + y)(x βˆ’ y) FACTORING POLYNOMIALS For the following exercises, find the greatest common factor. 45. 81p + 9pq βˆ’ 27p2q 2 46. 12x2y + 4xy 2 βˆ’18xy 47. 88a3b + 4a2b βˆ’ 144a2 For the following exercises, factor the polynomial. 48. 2x 2 βˆ’ 9x βˆ’ 18 51. x 2 + 10x + 25 54. 361x 2 βˆ’ 121 57. 64q 3 βˆ’ 27p3 1 2 _ _ 60. 4r (2r βˆ’ 1) βˆ’ βˆ’ 5(2r βˆ’ 1) 3 3 49. 8a2 + 30a βˆ’ 27 52. y 2 βˆ’ 6y + 9 55. p3 + 216 50. d 2 βˆ’ 5d βˆ’ 66 53. 4h2 βˆ’ 12hk + 9k2 56. 8x3 βˆ’ 125 3 1 _ _ 58. 4x(x βˆ’ 1) βˆ’ + 3(x βˆ’ 1) 4 4 4 1 _ _ βˆ’ 8(p + 3) 59. 3p(p + 3) 3 3 RATIONAL EXPRESSIONS For the following exercises, simplify the expression. 61. x2 βˆ’ x βˆ’ 12 ___________ x2 βˆ’ 8x + 16 62. 4y2 βˆ’ 25 _____________ 4y2 βˆ’ 20y + 25 63. 2a2 βˆ’ a βˆ’ 3 ___________ Β· 2a2 βˆ’ 6a βˆ’ 8 5a2 βˆ’ 19a βˆ’ 4 _____________ 10a2 βˆ’ 13a βˆ’ 3 65. m2 + 5m + 6 ____________ Γ· 2m2 βˆ’ 5m βˆ’ 3 2m2 + 3m βˆ’ 9 ____________ 4m2 βˆ’ 4m βˆ’ 3 67. 6 10 _ y _ x + 69. 1 2 _ c _ + d _________
6c + 12d _ dc 64. d βˆ’ 4 ______ Β· d 2 βˆ’ 9 d βˆ’ 3 _______ d 2 βˆ’ 16 66. 4d 2 βˆ’ 7d βˆ’ 2 _____________ 6d 2 βˆ’ 17d + 10 Γ· 8d 2 + 6d + 1 ____________ 6d 2 + 7d βˆ’ 10 68. 12 __________ βˆ’ a2 + 2a + 1 3 _____ a2 βˆ’1 70. 3 7 _ y _ x βˆ’ _______ 2 _ x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 7 2 CHAPTER 1 PREREQUISITES CHAPTER 1 PRACTICE TEST For the following exercises, identify the number as rational, irrational, whole, or natural. Choose the most descriptive answer. 1. βˆ’13 2. √ β€” 2 For the following exercises, evaluate the equations. 3. 2(x + 3) βˆ’ 12 = 18 4. y(3 + 3)2 βˆ’ 26 = 10 5. Write the number in standard notation: 3.1415 Γ— 106 6. Write the number in scientific notation: 0.0000000212. For the following exercises, simplify the expression. 7. βˆ’2 Β· (2 + 3 Β· 2)2 + 144 8. 4(x + 3) βˆ’ (6x + 2) 9. 35 Β· 3βˆ’3 3 2 ) 10. ( _ 3 13. √ β€” 441 11. 8x3 ____ (2x)2 14. √ β€” 490 12. (16y 0)2y βˆ’2 15. √ ___ 9x _ 16 16. β€” 121b2 √ _______ b 1 + √ β€” 17. 6 √ β€” 24 + 7 √ β€” 54 βˆ’ 12 √ β€” 6 18. β€” 3 βˆ’8 √ _ 625 √ β€” 4 19. (13q3 + 2q2 βˆ’ 3) βˆ’ (6q2 + 5q βˆ’ 3) 20. (6p2 + 2p + 1) + (9p2 βˆ’ 1) 21. (n βˆ’ 2)(n2 βˆ’ 4n + 4) 22. (a βˆ’ 2b)(2a + b) For the following exercises, factor the polynomial. 23. 16x 2 βˆ’ 81 24. y2 + 12y + 36 25. 27c 3 βˆ’ 1331 3 1 __ _ 26. 3x(x βˆ’ 6)
βˆ’ + 2(x βˆ’ 6) 4 4 For the following exercises, simplify the expression. 27. 2z2 + 7z + 3 ___________ Β· z2 βˆ’ 9 4z2 βˆ’ 15z + 9 ____________ 4z2 βˆ’ 1 x 2 _ x _ y + 28. 29. a 2b ___ ___ βˆ’ 9a 2b ________ 3a βˆ’ 2b _______ 6a Download the OpenStax text for free at http://cnx.org/content/col11759/latest. Equations and Inequalities 2 2 Figure 1 CHAPTER OUTLINE 2.1 The Rectangular Coordinate Systems and Graphs 2.2 Linear Equations in One Variable 2.3 Models and Applications 2.4 Complex Numbers 2.5 Quadratic Equations 2.6 Other Types of Equations 2.7 Linear Inequalities and Absolute Value Inequalities Introduction For most people, the term territorial possession indicates restrictions, usually dealing with trespassing or rite of passage and takes place in some foreign location. What most Americans do not realize is that from September through December, territorial possession dominates our lifestyles while watching the NFL. In this area, territorial possession is governed by the referees who make their decisions based on what the chains reveal. If the ball is at point A (x1, y1), then it is up to the quarterback to decide which route to point B (x2, y2), the end zone, is most feasible. 73 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 7 4 CHAPTER 2 EQUATIONS AND INEQUALITIES LEARNING OBJECTIVES In this section you will: β€’ Plot and identify ordered pairs in a Cartesian coordinate system. β€’ Graph equations by plotting ordered pairs. β€’ Find x-intercepts and y-intercepts of the graph of an equation. β€’ Use the distance formula. β€’ Use the midpoint formula. 2.1 THE RECTANGULAR COORDINATE SYSTEMS AND GRAPHS y Schiller Avenue Mannhelm Road Bertau Avenue McLean Street Wolf Road North Avenue x Figure 1 Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in Figure 1. Laying a rectangular coordinate grid over the map, we can see that each stop aligns with an intersection of grid lines.
In this section, we will learn how to use grid lines to describe locations and changes in locations. Plotting Ordered Pairs in the Cartesian Coordinate System An old story describes how seventeenth-century philosopher/mathematician RenΓ© Descartes invented the system that has become the foundation of algebra while sick in bed. According to the story, Descartes was staring at a fly crawling on the ceiling when he realized that he could describe the fly’s location in relation to the perpendicular lines formed by the adjacent walls of his room. He viewed the perpendicular lines as horizontal and vertical axes. Further, by dividing each axis into equal unit lengths, Descartes saw that it was possible to locate any object in a two-dimensional plane using just two numbersβ€”the displacement from the horizontal axis and the displacement from the vertical axis. While there is evidence that ideas similar to Descartes’ grid system existed centuries earlier, it was Descartes who introduced the components that comprise the Cartesian coordinate system, a grid system having perpendicular axes. Descartes named the horizontal axis the x-axis and the vertical axis the y-axis. The Cartesian coordinate system, also called the rectangular coordinate system, is based on a two-dimensional plane consisting of the x-axis and the y-axis. Perpendicular to each other, the axes divide the plane into four sections. Each section is called a quadrant; the quadrants are numbered counterclockwise as shown in Figure 2. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.1 THE RECTANGULAR COORDINATE SYSTEMS AND GRAPHS 75 y-axis Quadrant II Quadrant I x-axis Quadrant III Quadrant IV Figure 2 The center of the plane is the point at which the two axes cross. It is known as the origin, or point (0, 0). From the origin, each axis is further divided into equal units: increasing, positive numbers to the right on the x-axis and up the y-axis; decreasing, negative numbers to the left on the x-axis and down the y-axis. The axes extend to positive and negative infinity as shown by the arrowheads in Figure 3. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure 3 Each point in the plane is identified by its x
-coordinate, or horizontal displacement from the origin, and its y-coordinate, or vertical displacement from the origin. Together, we write them as an ordered pair indicating the combined distance from the origin in the form (x, y). An ordered pair is also known as a coordinate pair because it consists of x- and y-coordinates. For example, we can represent the point (3, βˆ’1) in the plane by moving three units to the right of the origin in the horizontal direction, and one unit down in the vertical direction. See Figure 4. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 x 21 4 (3, –1) 5 Figure 4 When dividing the axes into equally spaced increments, note that the x-axis may be considered separately from the y-axis. In other words, while the x-axis may be divided and labeled according to consecutive integers, the y-axis may be divided and labeled by increments of 2, or 10, or 100. In fact, the axes may represent other units, such as years against the balance in a savings account, or quantity against cost, and so on. Consider the rectangular coordinate system primarily as a method for showing the relationship between two quantities. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 7 6 CHAPTER 2 EQUATIONS AND INEQUALITIES Cartesian coordinate system A two-dimensional plane where the β€’ x-axis is the horizontal axis β€’ y-axis is the vertical axis A point in the plane is defined as an ordered pair, (x, y), such that x is determined by its horizontal distance from the origin and y is determined by its vertical distance from the origin. Example 1 Plotting Points in a Rectangular Coordinate System Plot the points (βˆ’2, 4), (3, 3), and (0, βˆ’3) in the plane. Solution To plot the point (βˆ’2, 4), begin at the origin. The x-coordinate is βˆ’2, so move two units to the left. The y-coordinate is 4, so then move four units up in the positive y direction. To plot the point (3, 3), begin again at the origin. The x-coordinate is 3, so move three units to the right. The y-coordinate is also 3, so move three units up in the positive y direction. To plot
the point (0, βˆ’3), begin again at the origin. The x-coordinate is 0. This tells us not to move in either direction along the x-axis. The y-coordinate is –3, so move three units down in the negative y direction. See the graph in Figure 5. (–2, 4) –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (3, 3) 321 4 5 x (0, –3) Figure 5 Analysis Note that when either coordinate is zero, the point must be on an axis. If the x-coordinate is zero, the point is on the y-axis. If the y-coordinate is zero, the point is on the x-axis. Graphing Equations by Plotting Points We can plot a set of points to represent an equation. When such an equation contains both an x variable and a y variable, it is called an equation in two variables. Its graph is called a graph in two variables. Any graph on a twodimensional plane is a graph in two variables. Suppose we want to graph the equation y = 2x βˆ’ 1. We can begin by substituting a value for x into the equation and determining the resulting value of y. Each pair of x- and y-values is an ordered pair that can be plotted. Table 1 lists values of x from βˆ’3 to 3 and the resulting values for y. x βˆ’3 βˆ’2 βˆ’1 0 1 2 3 y = 2x βˆ’ 1 y = 2(βˆ’3) βˆ’ 1 = βˆ’7 y = 2(βˆ’2) βˆ’ 1 = βˆ’5 y = 2(βˆ’1) βˆ’ 1 = βˆ’3 y = 2(0) βˆ’ 1 = βˆ’1 y = 2(1) βˆ’ 1 = 1 y = 2(2) βˆ’ 1 = 3 y = 2(3) βˆ’ 1 = 5 Table 1 (x, y) (βˆ’3, βˆ’7) (βˆ’2, βˆ’5) (βˆ’1, βˆ’3) (0, βˆ’1) (1, 1) (2, 3) (3, 5) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.1 THE RECTANGULAR COORDINATE SYSTEMS AND GRAPHS 77 We can plot the points in the table. The points for this particular equation form a line, so we
can connect them. See Figure 6. This is not true for all equations. (4, 7) (3, 5) (2, 3) (1, 1) 321 (0, –11 –2 –3 –4 –5 –6 –7 –5 –4 –3 –2 –1 (–1, –3) (–2, –5) (–3, –7) Figure 6 Note that the x-values chosen are arbitrary, regardless of the type of equation we are graphing. Of course, some situations may require particular values of x to be plotted in order to see a particular result. Otherwise, it is logical to choose values that can be calculated easily, and it is always a good idea to choose values that are both negative and positive. There is no rule dictating how many points to plot, although we need at least two to graph a line. Keep in mind, however, that the more points we plot, the more accurately we can sketch the graph. How To… Given an equation, graph by plotting points. 1. Make a table with one column labeled x, a second column labeled with the equation, and a third column listing the resulting ordered pairs. 2. Enter x-values down the first column using positive and negative values. Selecting the x-values in numerical order will make the graphing simpler. 3. Select x-values that will yield y-values with little effort, preferably ones that can be calculated mentally. 4. Plot the ordered pairs. 5. Connect the points if they form a line. Example 2 Graphing an Equation in Two Variables by Plotting Points Graph the equation y = βˆ’x + 2 by plotting points. Solution First, we construct a table similar to Table 2. Choose x values and calculate y. x βˆ’5 βˆ’3 βˆ’ = βˆ’ (βˆ’5) + 2 = 7 y = βˆ’ (βˆ’3) + 2 = 5 y = βˆ’ (βˆ’1) + 2 = 3 y = βˆ’ (0) + 2 = 2 y = βˆ’ (1) + 2 = 1 y = βˆ’ (3) + 2 = βˆ’1 y = βˆ’ (5) + 2 = βˆ’3 Table 2 (x, y) (βˆ’5, 7) (βˆ’3, 5) (βˆ’1, 3) (0, 2) (1, 1) (3, βˆ’1) (5, βˆ’3) Download the OpenStax text for free at http://cnx.org/content/col11759
/latest. 7 8 CHAPTER 2 EQUATIONS AND INEQUALITIES Now, plot the points. Connect them if they form a line. See Figure 7. (–5, 7) (–3, 5) (–1, 30, 2) (1, 1) –7 –6 –5 –4 –3 –2 –1 –1 –2 –3 321 (3, –1) 4 5 6 7 (5, –3) Figure 7 x Try It #1 1 _ x + 2. Construct a table and graph the equation by plotting points: y = 2 Graphing Equations with a Graphing Utility Most graphing calculators require similar techniques to graph an equation. The equations sometimes have to be manipulated so they are written in the style y =. The TI-84 Plus, and many other calculator makes and models, have a mode function, which allows the window (the screen for viewing the graph) to be altered so the pertinent parts of a graph can be seen. For example, the equation y = 2x βˆ’ 20 has been entered in the TI-84 Plus shown in Figure 8a. In Figure 8b, the resulting graph is shown. Notice that we cannot see on the screen where the graph crosses the axes. The standard window screen on the TI-84 Plus shows βˆ’10 ≀ x ≀ 10, and βˆ’10 ≀ y ≀ 10. See Figure 8c. Plot1 Plot2 Plot3 \Y1= 2X–20 \Y2= \Y3= \Y4= \Y5= \Y6= \Y7= (a) WINDOW Xmin = βˆ’10 Xmax = 10 Xscl = 1 Ymin = βˆ’10 Ymax = 10 Yscl = 1 Xres = 1 x (b) (c) Figure 8 (a) Enter the equation. (b) This is the graph in the original window. (c) These are the original settings. By changing the window to show more of the positive x-axis and more of the negative y-axis, we have a much better view of the graph and the x- and y-intercepts. See Figure 9a and Figure 9b. WINDOW Xmin = βˆ’5 Xmax = 15 Xscl = 1 Ymin = βˆ’30 Ymax = 10 Yscl = 1 Xres = 1 y (a) (b) x Figure 9 (a) This screen shows the new window settings. (b)
We can clearly view the intercepts in the new window. Example 3 Using a Graphing Utility to Graph an Equation 2 4 _ _ Use a graphing utility to graph the equation: y = βˆ’ x βˆ’. 3 3 Solution Enter the equation in the y = function of the calculator. Set the window settings so that both the x- and y-intercepts are showing in the window. See Figure 105 –4 –3 –2 –1 –1 –2 321 4 5 x Figure 10 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.1 THE RECTANGULAR COORDINATE SYSTEMS AND GRAPHS 79 Finding x-intercepts and y-intercepts The intercepts of a graph are points at which the graph crosses the axes. The x-intercept is the point at which the graph crosses the x-axis. At this point, the y-coordinate is zero. The y-intercept is the point at which the graph crosses the y-axis. At this point, the x-coordinate is zero. To determine the x-intercept, we set y equal to zero and solve for x. Similarly, to determine the y-intercept, we set x equal to zero and solve for y. For example, lets find the intercepts of the equation y = 3x βˆ’ 1. To find the x-intercept, set y = 0. To find the y-intercept, set x = 0. y = 3x βˆ’ 1 0 = 3x βˆ’ 1 1 = 3x = 3x βˆ’ 1 y = 3(0) βˆ’ 1 y = βˆ’1 (0, βˆ’1) x-intercept y-intercept We can confirm that our results make sense by observing a graph of the equation as in Figure 11. Notice that the graph crosses the axes where we predicted it would. –5 –4 –3 –2 y = 3x βˆ’ 1 321 4 5 x y 4 3 2 1 –1 –1 –2 –3 –4 Figure 11 given an equation, find the intercepts. β€’ Find the x-intercept by setting y = 0 and solving for x. β€’ Find the y-intercept by setting x = 0 and solving for y. Example 4 Finding the Intercepts of the Given Equation Find the intercepts of the equation y = βˆ’3x βˆ’ 4. Then sketch the graph using only the
intercepts. Solution Set y = 0 to find the x-intercept. Set x = 0 to find the y-intercept. y = βˆ’3x βˆ’ 4 0 = βˆ’3x βˆ’ 4 4 = βˆ’3x 3x βˆ’ 4 y = βˆ’3(0) βˆ’ 4 y = βˆ’4 (0, βˆ’4) xβˆ’intercept yβˆ’intercept Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 80 CHAPTER 2 EQUATIONS AND INEQUALITIES Plot both points, and draw a line passing through them as in Figure 12. 4 βˆ’ _, 0 3 –5 –4 –3 –2 y 3 2 1 –1 –1 –2 –3 –4 –5 –6 321 4 5 x (0, –4) Figure 12 Try It #2 3 _ x + 3. Find the intercepts of the equation and sketch the graph: y = βˆ’ 4 Using the Distance Formula Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem, a2 + b2 = c2, is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse. See Figure 13x1, y1) (x2, y2) d = c y2 βˆ’ y1 = b βˆ’ x1 = a x2 2 3 4 1 (x2, y1) 5 6 7 x Figure 13 The relationship of sides |x2 βˆ’ x1 | to side d is the same as that of sides a and b to side c. We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, |βˆ’3| = 3. ) The symbols |x2 βˆ’ x1 | indicate that the lengths of the sides of the triangle are positive. To find the length c, take the square root of both sides of the Pythagorean Theorem. | and |y2 βˆ’ y1 | and |y2 βˆ’ y1 It follows that the distance formula is given as c2 = a2 + b2 β†’ c = √ β€” a2 + b2 We do not have to use the absolute value symbols in this definition because any number squared is positive. d 2 = (x2 βˆ’ x1
)2 + (y2 βˆ’ y1)2 β†’ d = √ β€”β€” (x2 βˆ’ x1)2 + (y2 βˆ’ y1)2 the distance formula Given endpoints (x1, y1) and (x2, y2), the distance between two points is given by d = √ β€”β€” (x2 βˆ’ x1)2 + (y2 βˆ’ y1)2 Example 5 Finding the Distance between Two Points Find the distance between the points (βˆ’3, βˆ’1) and (2, 3). Solution Let us first look at the graph of the two points. Connect the points to form a right triangle as in Figure 14. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.1 THE RECTANGULAR COORDINATE SYSTEMS AND GRAPHS 81 y 4 3 2 1 –1 –1 –2 –3 –4 –2 –3 –4 –5 (βˆ’3, βˆ’1) (2, 3) x 321 4 5 (2, βˆ’1) Figure 14 Then, calculate the length of d using the distance formula. d = √ d = √ = √ β€”β€” β€”β€” (x2 βˆ’ x1)2 + (y2 βˆ’ y1)2 (2 βˆ’ (βˆ’3))2 + (3 βˆ’ (βˆ’1))2 (5)2 + (4)2 β€” = √ β€” 25 + 16 = √ β€” 41 Try It #3 Find the distance between two points: (1, 4) and (11, 9). Example 6 Finding the Distance Between Two Locations Let’s return to the situation introduced at the beginning of this section. Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in Figure 1. Find the total distance that Tracie traveled. Compare this with the distance between her starting and final positions. Solution The first thing we should do is identify ordered pairs to describe each position. If we set the starting position at the origin, we can identify each of the other points by counting units east (right) and north (up) on the grid. For example, the first stop is 1 block east and 1 block north, so it is at (1, 1). The next stop is 5 blocks to the east, so it is at (5
, 1). After that, she traveled 3 blocks east and 2 blocks north to (8, 3). Lastly, she traveled 4 blocks north to (8, 7). We can label these points on the grid as in Figure 15. y Schiller Avenue (8, 7) Mannhelm Road (8, 3) Wolf Road x Bertau Avenue McLean Street North Avenue (1, 1) (5, 1) (0, 0) Figure 15 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 82 CHAPTER 2 EQUATIONS AND INEQUALITIES Next, we can calculate the distance. Note that each grid unit represents 1,000 feet. β€’ From her starting location to her first stop at (1, 1), Tracie might have driven north 1,000 feet and then east 1,000 feet, or vice versa. Either way, she drove 2,000 feet to her first stop. β€’ Her second stop is at (5, 1). So from (1, 1) to (5, 1), Tracie drove east 4,000 feet. β€’ Her third stop is at (8, 3). There are a number of routes from (5, 1) to (8, 3). Whatever route Tracie decided to use, the distance is the same, as there are no angular streets between the two points. Let’s say she drove east 3,000 feet and then north 2,000 feet for a total of 5,000 feet. β€’ Tracie’s final stop is at (8, 7). This is a straight drive north from (8, 3) for a total of 4,000 feet. Next, we will add the distances listed in Table 3. From/To Number of Feet Driven (0, 0) to (1, 1) (1, 1) to (5, 1) (5, 1) to (8, 3) (8, 3) to (8, 7) Total 2,000 4,000 5,000 4,000 15,000 Table 3 The total distance Tracie drove is 15,000 feet, or 2.84 miles. This is not, however, the actual distance between her starting and ending positions. To find this distance, we can use the distance formula between the points (0, 0) and (8, 7). d = √ β€” (8 βˆ’ 0)2 + (7 βˆ’ 0
)2 = √ β€” 64 + 49 = √ β€” 113 β‰ˆ 10.63 units At 1,000 feet per grid unit, the distance between Elmhurst, IL, to Franklin Park is 10,630.14 feet, or 2.01 miles. The distance formula results in a shorter calculation because it is based on the hypotenuse of a right triangle, a straight diagonal from the origin to the point (8, 7). Perhaps you have heard the saying β€œas the crow flies,” which means the shortest distance between two points because a crow can fly in a straight line even though a person on the ground has to travel a longer distance on existing roadways. Using the Midpoint Formula When the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the midpoint formula. Given the endpoints of a line segment, (x1, y1) and (x2, y2), the midpoint formula states how to find the coordinates of the midpoint M. x1 + x2 _______, 2 A graphical view of a midpoint is shown in Figure 16. Notice that the line segments on either side of the midpoint are congruent. y1 + y2 ) _______ 2 M = ( y 0 x1 + x2 _______, 2 y1 + y2 _______ 2 x Figure 16 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.1 THE RECTANGULAR COORDINATE SYSTEMS AND GRAPHS 83 Example 7 Finding the Midpoint of the Line Segment Find the midpoint of the line segment with the endpoints (7, βˆ’2) and (9, 5). Solution Use the formula to find the midpoint of the line segment. x1 + x2 ( _______, 2 y1 + y2 _______ 2 ) = ( βˆ’2 + 5 7 + 9 ) _______ _____, 2 2 3 ) = ( 8, __ 2 Try It #4 Find the midpoint of the line segment with endpoints (βˆ’2, βˆ’1) and (βˆ’8, 6). Example 8 Finding the Center of a Circle The diameter of a circle has endpoints (βˆ’1, βˆ’4) and (5, βˆ’4). Find the center of the circle. Solution The center of a circle is the center, or midpoint,
of its diameter. Thus, the midpoint formula will yield the center point. x1 + x2 ( _______, 2 y1 + y2 ) _______ 2 βˆ’4 βˆ’4 βˆ’1 + 5 ( ______ _______, 2 2 8 4 ) = (2, βˆ’4) ) = ( __ __, βˆ’ 2 2 Access these online resources for additional instruction and practice with the Cartesian coordinate system. β€’ Plotting Points on the Coordinate Plane (http://openstaxcollege.org/l/coordplotpnts) β€’ Find x- and y-intercepts Based on the Graph of a Line (http://openstaxcollege.org/l/xyintsgraph) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 84 CHAPTER 2 EQUATIONS AND INEQUALITIES 2.1 SECTION EXERCISES VERBAL 1. Is it possible for a point plotted in the Cartesian coordinate system to not lie in one of the four quadrants? Explain. 3. Describe in your own words what the y-intercept of a graph is. 2. Describe the process for finding the x-intercept and the y-intercept of a graph algebraically. 4. When using the distance formula β€”β€” (x2 βˆ’ x1)2 + (y2 βˆ’ y1)2, explain the correct d = √ order of operations that are to be performed to obtain the correct answer. ALGEBRAIC For each of the following exercises, find the x-intercept and the y-intercept without graphing. Write the coordinates of each intercept. 5. y = βˆ’3x + 6 7. 3x βˆ’ 2y = 6 8. 4x βˆ’ 3 = 2y 9. 3x + 8y = 9 6. 4y = 2x βˆ’ 1 3 2 _ _ 10. 2x βˆ’ y + 3 = 4 3 For each of the following exercises, solve the equation for y in terms of x. 11. 4x + 2y = 8 15. 5y + 4 = 10x 12. 3x βˆ’ 2y = 6 16. 5x + 2y = 0 13. 2x = 5 βˆ’ 3y 14. x βˆ’ 2y = 7 For each of the following exercises, find the distance between the two points. Simplify your answers, and write the exact answer in simplest radical form for irrational answers. 17
. (βˆ’4, 1) and (3, βˆ’4) 20. (βˆ’4, 3) and (10, 3) 18. (2, βˆ’5) and (7, 4) 19. (5, 0) and (5, 6) 21. Find the distance between the two points given using your calculator, and round your answer to the nearest hundredth. (19, 12) and (41, 71) For each of the following exercises, find the coordinates of the midpoint of the line segment that joins the two given points. 22. (βˆ’5, βˆ’6) and (4, 2) 24. (βˆ’5, βˆ’3) and (βˆ’2, βˆ’8) 23. (βˆ’1, 1) and (7, βˆ’4) 25. (0, 7) and (4, βˆ’9) 26. (βˆ’43, 17) and (23, βˆ’34) GRAPHICAL For each of the following exercises, identify the information requested. 27. What are the coordinates of the origin? 28. If a point is located on the y-axis, what is the x-coordinate? 29. If a point is located on the x-axis, what is the y-coordinate? For each of the following exercises, plot the three points on the given coordinate plane. State whether the three points you plotted appear to be collinear (on the same line). 30. (4, 1)(βˆ’2, βˆ’3)(5, 0) 31. (βˆ’1, 2)(0, 4)(2, 1) y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.1 SECTION EXERCISES 85 32. (βˆ’3, 0)(βˆ’3, 4)(βˆ’3, βˆ’3) 33. Name the coordinates of the points graphed. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x A –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 B C 5 x 21 3 4 34.
Name the quadrant in which the following points would be located. If the point is on an axis, name the axis. a. (βˆ’3, βˆ’4) b. (βˆ’5, 0) c. (1, βˆ’4) d. (βˆ’2, 7) e. (0, βˆ’3) For each of the following exercises, construct a table and graph the equation by plotting at least three points. 1 _ 37. 2y = x + 3 35. y = x + 2 3 36. y = βˆ’3x + 1 NUMERIC For each of the following exercises, find and plot the x- and y-intercepts, and graph the straight line based on those two points. 38. 4x βˆ’ 3y = 12 41. 3y = βˆ’2x + 6 39. x βˆ’ 2y = 8 40. y βˆ’ 5 = 5x 42. y = x βˆ’ 3 _____ 2 For each of the following exercises, use the graph in the figure below. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x 43. Find the distance between the two endpoints using the distance formula. Round to three decimal places. 44. Find the coordinates of the midpoint of the line segment connecting the two points. 45. Find the distance that (βˆ’3, 4) is from the origin. 46. Find the distance that (5, 2) is from the origin. Round to three decimal places. 47. Which point is closer to the origin? TECHNOLOGY For the following exercises, use your graphing calculator to input the linear graphs in the Y= graph menu. After graphing it, use the 2nd CALC button and 1:value button, hit ENTER. At the lower part of the screen you will see β€œx=” and a blinking cursor. You may enter any number for x and it will display the y value for any x value you input. Use this and plug in x = 0, thus finding the y-intercept, for each of the following graphs. 48. Y1 = βˆ’2x + 5 49. Y1 = 3x βˆ’ 8 ______ 4 50. Y1 = x + 5 _____ 2 For the following exercises, use your graphing calculator to input the linear graphs in the Y= graph menu. After graphing it, use the 2nd CALC button and 2:zero button, hit ENTER. At
the lower part of the screen you will see β€œleft bound?” and a blinking cursor on the graph of the line. Move this cursor to the left of the x-intercept, hit ENTER. Now it says β€œright bound?” Move the cursor to the right of the x-intercept, hit ENTER. Now it says β€œguess?” Move your cursor to the left somewhere in between the left and right bound near the x-intercept. Hit ENTER. At the bottom of your screen it will display the coordinates of the x-intercept or the β€œzero” to the y-value. Use this to find the x-intercept. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 86 CHAPTER 2 EQUATIONS AND INEQUALITIES Note: With linear/straight line functions the zero is not really a β€œguess,” but it is necessary to enter a β€œguess” so it will search and find the exact x-intercept between your right and left boundaries. With other types of functions (more than one x-intercept), they may be irrational numbers so β€œguess” is more appropriate to give it the correct limits to find a very close approximation between the left and right boundaries. 51. Y1 = βˆ’8x + 6 52. Y1 = 4x βˆ’ 7 53. Y1 = 3x + 5 ______ 4 Round your answer to the nearest thousandth. EXTENSIONS 54. A man drove 10 mi directly east from his home, made a left turn at an intersection, and then traveled 5 mi north to his place of work. If a road was made directly from his home to his place of work, what would its distance be to the nearest tenth of a mile? 55. If the road was made in the previous exercise, how much shorter would the man’s one-way trip be every day? 56. Given these four points: A(1, 3), B(βˆ’3, 5), C(4, 7), and D(5, βˆ’4), find the coordinates of the midpoint _ #CD. of line segments _ #AB and 57. After finding the two midpoints in the previous exercise, find the distance between the two midpoints to the nearest thousandth. 58. Given the graph of the rectangle shown and the 59. In the previous exercise, find the
coordinates of the coordinates of its vertices, prove that the diagonals of the rectangle are of equal length. midpoint for each diagonal. (–6, 5) y 6 5 4 3 2 1 (10, 5) –5 –7 –4 (–6, –1) –3 –2 –1 –1 –2 –3 321 4 5 6 7 8 9 11 (10, –1) x REAL-WORLD APPLICATIONS 60. The coordinates on a map for San Francisco are (53, 17) and those for Sacramento are (123, 78). Note that coordinates represent miles. Find the distance between the cities to the nearest mile. 61. If San Jose’s coordinates are (76, βˆ’12), where the coordinates represent miles, find the distance between San Jose and San Francisco to the nearest mile. 62. A small craft in Lake Ontario sends out a distress signal. The coordinates of the boat in trouble were (49, 64). One rescue boat is at the coordinates (60, 82) and a second Coast Guard craft is at coordinates (58, 47). Assuming both rescue craft travel at the same rate, which one would get to the distressed boat the fastest? 64. If we rent a truck and pay a $75/day fee plus $.20 for every mile we travel, write a linear equation that would express the total cost y, using x to represent the number of miles we travel. Graph this function on your graphing calculator and find the total cost for one day if we travel 70 mi. 63. A man on the top of a building wants to have a guy wire extend to a point on the ground 20 ft from the building. To the nearest foot, how long will the wire have to be if the building is 50 ft tall? (20, 50) 50 (0, 0) 20 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 LINEAR EQUATIONS IN ONE VARIABLE 87 LEARNING OBJECTIVES In this section you will: β€’ ole linear equations in one ariable. β€’ ole rational equations in one ariable. β€’ Find the slope of a line that contains two gien points. β€’ Graph linear equations. β€’ Use the point-slope andor slope intercept formula to write the equation of a line that satisfies certain gien properties. β€’ Gien the equations of two lines determine whether their graphs are parallel perpendicular
or neither. β€’ Find the equation of a line parallel or perpendicular to a gien line. 2.2 LINEAR EQUATIONS IN ONE VARIABLE Figure 1 y Figure 1 x Solving Linear Equations in One Variable linear equationTh ax+b = identity equation x =x +x solution set x conditional equation x +=x βˆ’ x +=x βˆ’ x =βˆ’ x =βˆ’ Thβˆ’ inconsistent equationx βˆ’=xβˆ’ x βˆ’=x βˆ’ x βˆ’βˆ’x =x βˆ’βˆ’x x βˆ’β‰ βˆ’ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 88 CHAPTER 2 EQUATIONS AND INEQUALITIES Indeed, βˆ’15 β‰  βˆ’20. There is no solution because this is an inconsistent equation. Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows. linear equation in one variable A linear equation in one variable can be written in the form ax + b = 0 where a and b are real numbers, a β‰  0. How To… Given a linear equation in one variable, use algebra to solve it. The following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads x =, if x is the unknown. There is no set order, as the steps used depend on what is given: 1. We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero. 2. Apply the distributive property as needed: a(b + c) = ab + ac. 3. Isolate the variable on one side of the equation. 4. When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient. Example 1 Solving an Equation in One Variable Solve the following equation: 2x + 7 = 19. Solution This equation can be written in the form ax + b = 0 by subtracting 19 from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations. 2x + 7 = 19 2x = 12 x = 6 Subtract 7 from both sides. 1 _ or divide by 2. Multiply both sides by 2 The solution is 6. Try It #1 Solve the linear equation in one variable:
2x + 1 = βˆ’9. Example 2 Solving an Equation Algebraically When the Variable Appears on Both Sides Solve the following equation: 4(x βˆ’ 3) + 12 = 15 βˆ’ 5(x + 6). Solution Apply standard algebraic properties. 4(x βˆ’ 3) + 12 = 15 βˆ’ 5(x + 6) 4x βˆ’ 12 + 12 = 15 βˆ’ 5x βˆ’ 30 Apply the distributive property. 4x = βˆ’15 βˆ’ 5x Combine like terms. x = βˆ’ 9x = βˆ’15 15_ 9 5 _ x = βˆ’ 3 Place x- terms on one side and simplify. 1 _, the reciprocal of 9. Multiply both sides by 9 Analysis This problem requires the distributive property to be applied twice, and then the properties of algebra are used 5 _ to reach the final line, x = βˆ’. 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 LINEAR EQUATIONS IN ONE VARIABLE 89 Try It #2 Solve the equation in one variable: βˆ’2(3x βˆ’ 1) + x = 14 βˆ’ x. Solving a Rational Equation In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation. 7 2 _ _ Recall that a rational number is the ratio of two numbers, such as. A rational expression is the ratio, or quotient, or 2 3 of two polynomials. Here are three examples. x + 1 _, x2 βˆ’ 4 1 _ x βˆ’ 3, or 4 _ x2 + x βˆ’ 2 Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD). Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out. Example 3 Solving a Rational Equation Solve the rational equation: 7 _ 2x βˆ’ = 5 _ 3x 22 _. 3 Solution We have three denominato
rs; 2x, 3x, and 3. The LCD must contain 2x, 3x, and 3. An LCD of 6x contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD 6x. βˆ’ 7 5 22 ) = ( (6x) ( ) (6x) _ _ _ 2x 3x 3 5 7 22 ) (6x) ) = ( ) βˆ’ (6x) ( (6x) ( _ _ _ 3 3x 2x ) (, 6 x) ) = ( ) βˆ’ (, 6x ) ( (, 6x ) ( 7 5 22 _ _ _, 2x, 3x, 3 3(7) βˆ’ 2(5) = 22(2x) 21 βˆ’ 10 = 44x 11 = 44x = x 11 _ 44 1 _ = x 4 Use the distributive property. Cancel out the common factors. Multiply remaining factors by each numerator. A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomialβ€”two terms added or subtractedβ€”such as (x + 1). Always consider a binomial as an individual factorβ€”the terms cannot be separated. For example, suppose a problem has three terms and the denominators are x, x βˆ’ 1, and 3x βˆ’ 3. First, factor all denominators. We then have x, (x βˆ’ 1), and 3(x βˆ’ 1) as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of (x βˆ’ 1). The x in the first denominator is separate from the x in the (x βˆ’ 1) denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the x, one factor of (x βˆ’ 1), and the 3. Thus, the LCD is the following: x(x βˆ’ 1)3 = 3x(x βˆ’ 1) So, both sides of the equation would be multiplied by 3x(x βˆ’ 1). Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out. Download the OpenStax text for free at http://cnx.org/
content/col11759/latest. 90 CHAPTER 2 EQUATIONS AND INEQUALITIES Another example is a problem with two denominators, such as x and x 2 + 2x. Once the second denominator is factored as x 2 + 2x = x(x + 2), there is a common factor of x in both denominators and the LCD is x(x + 2). Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation. We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign = =., then If d b d b Multiply a(d) and b(c), which results in ad = bc. Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities. rational equations A rational equation contains at least one rational expression where the variable appears in at least one of the denominators. How To… Given a rational equation, solve it. 1. Factor all denominators in the equation. 2. Find and exclude values that set each denominator equal to zero. 3. Find the LCD. 4. Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left. 5. Solve the remaining equation. 6. Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator. Example 4 Solving a Rational Equation without Factoring Solve the following rational equation 2x Solution We have three denominators: x, 2, and 2x. No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is 2x. Only one value is excluded from a solution set, 0. Next, multiply the whole equation (both sides of the equal sign) by 2x. 7 3 2 2x ( ) = ( _ _ _ x βˆ’ 2x 2_ _ % 2 %x 2(2) βˆ’ 3x = 7 ) 2x ) %2x 7 _ % 2x 4 βˆ’ 3x = 7 βˆ’3x = 3 x = βˆ’1 or {βˆ’1} Distribute 2x. Denominators cancel out. The proposed solution is βˆ’1, which is not an excluded value, so the solution set contains one number, βˆ’1, or {βˆ’
1} written in set notation. Try It #3 Solve the rational equation: 2 _ 3x 1 _ βˆ’ = 4 1 _. 6x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 LINEAR EQUATIONS IN ONE VARIABLE 91 Example 5 Solving a Rational Equation by Factoring the Denominator 1 1 _ _ x = Solve the following rational equation: 10 βˆ’ 3 _. 4x Solution First find the common denominator. The three denominators in factored form are x, 10 = 2 Δ‹ 5, and 4x = 2 Δ‹ 2 Δ‹ x. The smallest expression that is divisible by each one of the denominators is 20x. Only x = 0 is an excluded value. Multiply the whole equation by 20x. 1 20x ( βˆ’ _ x ) = ( 1 _ 10 20 = 2x βˆ’ 15 ) 20x 3 ___ 4x The solution is 35 _. 2 35 = 2x 35 ___ 2 = x Try It #4 Solve the rational equation: βˆ’ 5 _ 2x + 3 _ 4x 7 _ = βˆ’. 4 Example 6 Solving Rational Equations with a Binomial in the Denominator Solve the following rational equations and state the excluded values: a. Solution. The denominators x and x βˆ’ 6 have nothing in common. Therefore, the LCD is the product x(x βˆ’ 6). However, for this problem, we can cross-multiply 3x = 5(x βˆ’ 6) 3x = 5x βˆ’ 30 βˆ’2x = βˆ’30 x = 15 Distribute. The solution is 15. The excluded values are 6 and 0. b. The LCD is 2(x βˆ’ 3). Multiply both sides of the equation by 2(x βˆ’ 3). 2(x βˆ’ 3) ( ) = ( x _____ x βˆ’ 3 2, (x βˆ’ 3) x ________ =, x βˆ’ 3 5 _____ x βˆ’ 3 2,(x βˆ’ 3) 5 ________ βˆ’, x βˆ’ 3 1 ) 2(x βˆ’ 3) __ βˆ’ 2 % 2 (x βˆ’ 3) _______ % 2 2x = 10 βˆ’ (x βˆ’ 3) 2x = 10 βˆ’ x + 3 2x = 13 βˆ’ x 3x = 13 13___ 3 x = The solution is. The excluded value is 3. 13 _ 3 Download the OpenStax
text for free at http://cnx.org/content/col11759/latest. 92 CHAPTER 2 EQUATIONS AND INEQUALITIES c. The least common denominator is 2(x βˆ’ 2). Multiply both sides of the equation by x(x βˆ’ 2). 2(x βˆ’ 2(x βˆ’ 2) _ βˆ’ 2 2x = 10 βˆ’ (x βˆ’ 2) 2x = 12 βˆ’ x 3x = 12 x = 4 The solution is 4. The excluded value is 2. Try It #5 Solve βˆ’3 _ 2x + 1 = 4 _ 3x + 1. State the excluded values. Example 7 Solving a Rational Equation with Factored Denominators and Stating Excluded Values Solve the rational equation after factoring the denominators:. State the excluded values = 2x _ x2 βˆ’ 1 Solution We must factor the denominator x2 βˆ’ 1. We recognize this as the difference of squares, and factor it as (x βˆ’ 1)(x + 1). Thus, the LCD that contains each denominator is (x βˆ’ 1)(x + 1). Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation. (x βˆ’ 1)(x + 1) ( ) (x βˆ’ 1)(x + 1) 2x __ (x βˆ’ 1)(x + 1(x βˆ’ 1) βˆ’ 1(x + 1) = 2x 2x βˆ’ 2 βˆ’ x βˆ’ 1 = 2x Distribute the negative sign. βˆ’3 βˆ’ x = 0 βˆ’3 = x The solution is βˆ’3. The excluded values are 1 and βˆ’1. Try It #6 Solve the rational equation. x2 βˆ’ x βˆ’ 2 Finding a Linear Equation Perhaps the most familiar form of a linear equation is the slope-intercept form, written as y = mx + b, where m = slope and b = y-intercept. Let us begin with the slope. The Slope of a Line The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run. m = y2 βˆ’ y1 __ x2 βˆ’ x1 If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As
the slope increases, the line becomes steeper. Some examples are shown in Figure 2. The lines indicate the following slopes: m = βˆ’3, 1 _ m = 2, and m =. 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 LINEAR EQUATIONS IN ONE VARIABLE 93 y y = 2x + 1 1 y = x + 2 3 42 6 8 10 12 x y = βˆ’3x βˆ’ 2 12 10 8 6 4 2 –12 –10 –8 –6 –4 –2 –2 –4 –6 –8 –10 –12 Figure 2 the slope of a line The slope of a line, m, represents the change in y over the change in x. Given two points, (x1, y1) and (x2, y2), the following formula determines the slope of a line containing these points: m = y2 βˆ’ y1 __ x2 βˆ’ x1 Example 8 Finding the Slope of a Line Given Two Points Find the slope of a line that passes through the points (2, βˆ’1) and (βˆ’5, 3). Solution We substitute the y-values and the x-values into the formula. m = 3 βˆ’ (βˆ’1) _ βˆ’5 βˆ’ 2 4 _ βˆ’7 4 _ = βˆ’ 7 = 4 _ The slope is βˆ’. 7 Analysis terms and the order of the x terms in the numerator and denominator, the calculation will yield the same result. It does not matter which point is called (x1, y1) or (x2, y2). As long as we are consistent with the order of the y Try It #7 Find the slope of the line that passes through the points (βˆ’2, 6) and (1, 4). Example 9 Identifying the Slope and y-intercept of a Line Given an Equation 3 _ Identify the slope and y-intercept, given the equation y = βˆ’ x βˆ’ 4. 4 3 _ Solution As the line is in y = mx + b form, the given line has a slope of m = βˆ’. The y-intercept is b = βˆ’4. 4 Analysis The y-intercept is the point at which the line crosses the y-axis. On the y-axis, x = 0. We can always identify the y-intercept when the line is in slope-intercept form, as it will always equal b
. Or, just substitute x = 0 and solve for y. The Point-Slope Formula Given the slope and one point on a line, we can find the equation of the line using the point-slope formula. y βˆ’ y1 = m(x βˆ’ x1) This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 94 CHAPTER 2 EQUATIONS AND INEQUALITIES the point-slope formula Given one point and the slope, the point-slope formula will lead to the equation of a line: y βˆ’ y1 = m(x βˆ’ x1) Example 10 Finding the Equation of a Line Given the Slope and One Point Write the equation of the line with slope m = βˆ’3 and passing through the point (4, 8). Write the final equation in slope-intercept form. Solution Using the point-slope formula, substitute βˆ’3 for m and the point (4, 8) for (x1, y1). y βˆ’ y1 = m(x βˆ’ x1) y βˆ’ 8 = βˆ’3(x βˆ’ 4) y βˆ’ 8 = βˆ’3x + 12 y = βˆ’3x + 20 Analysis Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained. Try It #8 Given m = 4, find the equation of the line in slope-intercept form passing through the point (2, 5). Example 11 Finding the Equation of a Line Passing Through Two Given Points Find the equation of the line passing through the points (3, 4) and (0, βˆ’3). Write the final equation in slope-intercept form. Solution First, we calculate the slope using the slope formula and two points. m = = βˆ’3 βˆ’ 4_ 0 βˆ’ 3 βˆ’7 _ βˆ’3 7 _ = 3 7 _ Next, we use the point-slope formula with the slope of, and either point. Let’s pick the point (3, 4) for (x1, y1). 3 7
_ (x βˆ’ 3. Distribute the 3 7 _ x βˆ’ 3. In slope-intercept form, the equation is written as y = 3 Analysis To prove that either point can be used, let us use the second point (0, βˆ’3) and see if we get the same equation. 7 _ y βˆ’ (βˆ’3) = (x βˆ’ 0 We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 LINEAR EQUATIONS IN ONE VARIABLE 95 Standard Form of a Line Another way that we can represent the equation of a line is in standard form. Standard form is given as Ax + By = C where A, B, and C are integers. The x- and y-terms are on one side of the equal sign and the constant term is on the other side. Example 12 Finding the Equation of a Line and Writing It in Standard Form 1, βˆ’2 ). Write the equation in standard form. Find the equation of the line with m = βˆ’6 and passing through the point ( _ 4 Solution We begin using the point-slope formula. 1 ) y βˆ’ (βˆ’2) = βˆ’6x + 2 From here, we multiply through by 2, as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right. 3 ) 2 2(y + 2) = ( βˆ’6x + _ 2 2y + 4 = βˆ’12x + 3 This equation is now written in standard form. 12x + 2y = βˆ’1 Try It #9 1 1 ). and passing through the point ( 1, _ _ Find the equation of the line in standard form with slope m = βˆ’ 3 3 Vertical and Horizontal Lines The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as x = c where c is a constant. The slope of a vertical line is undefined, and regardless of the y-value of any point on the line, the x-coordinate of the point will be c. Suppose that we want to find the equation of a line containing the following points: (βˆ’3, βˆ’5), (βˆ’3, 1), (βˆ’3,
3), and (βˆ’3, 5). First, we will find the slope. m = 5 βˆ’ 3 2 _________ __ = βˆ’3 βˆ’ (βˆ’3) 0 Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x-coordinates are the same and we find a vertical line through x = βˆ’3. See Figure 3. The equation of a horizontal line is given as y = c where c is a constant. The slope of a horizontal line is zero, and for any x-value of a point on the line, the y-coordinate will be c. Suppose we want to find the equation of a line that contains the following set of points: (βˆ’2, βˆ’2), (0, βˆ’2), (3, βˆ’2), and (5, βˆ’2). We can use the point-slope formula. First, we find the slope using any two points on the line. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 96 CHAPTER 2 EQUATIONS AND INEQUALITIES m = βˆ’2 βˆ’ (βˆ’2)__ 0 βˆ’ (βˆ’2) 0 _ = 2 = 0 Use any point for (x1, y1) in the formula, or use the y-intercept. y βˆ’ (βˆ’2) = 0(x βˆ’ 3) y + 2 = 0 y = βˆ’2 The graph is a horizontal line through y = βˆ’2. Notice that all of the y-coordinates are the same. See Figure 3. x = βˆ’3 –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 x 21 3 4 5 y = βˆ’2 Figure 3 The line x = βˆ’3 is a vertical line. The line y = βˆ’2 is a horizontal line. Example 13 Finding the Equation of a Line Passing Through the Given Points Find the equation of the line passing through the given points: (1, βˆ’3) and (1, 4). Solution The x-coordinate of both points is 1. Therefore, we have a vertical line, x = 1. Try It #10 Find the equation of the line passing through (βˆ’5, 2) and (2, 2). Determining Whether Graphs of Lines are Parallel or Perpendicular Parallel lines have the same slope and different y
-intercepts. Lines that are parallel to each other will never intersect. For example, Figure 4 shows the graphs of various lines with the same slope, m = 2. y = 2x βˆ’ 3 y = 2x + 1 y = 2x + 5 21 1 –1 –2 –3 –4 –5 –5 –4 –3 –2 Figure 4 Parallel lines All of the lines shown in the graph are parallel because they have the same slope and different y-intercepts. Lines that are perpendicular intersect to form a 90Β° -angle. The slope of one line is the negative reciprocal of the other. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 LINEAR EQUATIONS IN ONE VARIABLE 97 We can show that two lines are perpendicular if the product of the two slopes is βˆ’1: m1 Δ‹ m2 = βˆ’1. For example, Figure 5 shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of βˆ’ 1 __. 3 m1 Δ‹ m2 = βˆ’1 1 ) = βˆ’1 3 Δ‹ ( βˆ’ _ 3 y y = 3x βˆ’ 1 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 Figure 5 Perpendicular lines Example 14 Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: 3y = βˆ’ 4x + 3 and 3x βˆ’ 4y = 8. Solution The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form. First equation: Second equation: See the graph of both lines in Figure 6. 3y = βˆ’4x + 3x βˆ’ 4y = 8 βˆ’4y = βˆ’3x + 5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 6 From the graph, we can see that the lines appear perpendicular, but we must compare the slopes. 4 _ m1 = βˆ’ 3 3 _ m2 = 4 3 4 ) ( m1 Δ‹ m2 = ( βˆ’ ) = βˆ’1 _ _ 4 3 The slopes are negative reciprocals of each other, confirming that the
lines are perpendicular. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 98 CHAPTER 2 EQUATIONS AND INEQUALITIES Try It #11 Graph the two lines and determine whether they are parallel, perpendicular, or neither: 2y βˆ’ x = 10 and 2y = x + 4. Writing the Equations of Lines Parallel or Perpendicular to a Given Line As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line. How To… Given an equation for a line, write the equation of a line parallel or perpendicular to it. 1. Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form. 2. Use the slope and the given point with the point-slope formula. 3. Simplify the line to slope-intercept form and compare the equation to the given line. Example 15 Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point Write the equation of line parallel to a 5x + 3y = 1 and passing through the point (3, 5). Solution First, we will write the equation in slope-intercept form to find the slope. 5x + 3y = 1 3y = 5x + The slope is m = βˆ’, but that really does not enter into our problem, as the only thing we need. The y-intercept is 3 3 for two lines to be parallel is the same slope. The one exception is that if the y-intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula. 5 _ y βˆ’ 5 = βˆ’ (x βˆ’ 3 + 10 y = βˆ’ 3 5 _ The equation of the line is y = βˆ’ x + 10. See Figure 7. 3 y = βˆ’ x + + 10 3 –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 42 6 8 10 x Figure 7 Try It #12 Find the equation of the line parallel to 5x = 7 + y and passing through
the point (βˆ’1, βˆ’2). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 LINEAR EQUATIONS IN ONE VARIABLE 99 Example 16 Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point Find the equation of the line perpendicular to 5x βˆ’ 3y + 4 = 0 and passes through the point (βˆ’4, 1). Solution The first step is to write the equation in slope-intercept form. 5x βˆ’ 3y + 4 = 0 βˆ’3y = βˆ’5x βˆ’ 4 5 _ We see that the slope is m =. This means that the slope of the line perpendicular to the given line is the negative 3 3 _ reciprocal, or βˆ’. Next, we use the point-slope formula with this new slope and the given pointx βˆ’ (βˆ’4)) 5 3 12_ 12 Access these online resources for additional instruction and practice with linear equations. β€’ Solving rational equations (http://openstaxcollege.org/l/rationaleqs) β€’ Equation of a line given two points (http://openstaxcollege.org/l/twopointsline) β€’ Finding the equation of a line perpendicular to another line through a given point (http://openstaxcollege.org/l/findperpline) β€’ Finding the equation of a line parallel to another line through a given point (http://openstaxcollege.org/l/findparaline) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 100 CHAPTER 2 EQUATIONS AND INEQUALITIES 2.2 SECTION EXERCISES VERBAL 1. 2. 3. y =x + 5. x βˆ’ = x + 4. x =x =βˆ’ ALGEBRAIC x 6. x +=x βˆ’ 7. xβˆ’= 8. x +βˆ’=x + 9. βˆ’x +=x βˆ’ 10. βˆ’ x= 11. x = βˆ’ x + 12. = x + 13. x βˆ’+x =x + 14. x x + = βˆ’ 15. x + βˆ’ x βˆ’ = x x- 16. = x βˆ’ 17. βˆ’ x+ = x+ x+ 18. x βˆ’ = x βˆ’ + x βˆ’x βˆ’ 19. x x βˆ’ += x βˆ’ 20. x+ + xβˆ’
= βˆ’ xβˆ’xβˆ’ x = + 21. x 22. 23. βˆ’βˆ’ 24. βˆ’ 25. βˆ’βˆ’ 26. βˆ’ 27. 30. 33. 36. 39. 28. 1 3 31. 34. 37. 40. 29. 32. 35. 38. 41. 42. 44. 46. βˆ’ βˆ’βˆ’ 43. βˆ’ 45. βˆ’ 47. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 SECTION EXERCISES 101 48. βˆ’βˆ’ 49. 50. 52. (,) 51. βˆ’ 53. βˆ’βˆ’ GRAPHICAL 54. y =x + x βˆ’ y=βˆ’ NUMERIC 55. x βˆ’y = y βˆ’x = 56. x + y = y =x + 57. x = y =βˆ’ 58. βˆ’ βˆ’ 59. βˆ’βˆ’ 60. y =x + 61. y =x βˆ’ βˆ’ 62. = (,) + 63. = (,) + 64. = (,) 65. = (,) EXTENSIONS 66. y βˆ’y=mx βˆ’x x xyym 68. x βˆ’y = 70. REAL-WORLD APPLICATIONS 71. Th s 2.5 ft, fin x 67. Ax +By =C y ABCx. Th 69. βˆ’ 72. fi x y p =x +y y p = x = Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 102 CHAPTER 2 EQUATIONS AND INEQUALITIES LEARNING OBJECTIVES In this section you will: β€’ Set up a linear equation to solve a real-world application. 2.3 MODELS AND APPLICATIONS Figure 1 Credit: Kevin Dooley Josh is hoping to get an A in his college algebra class. He has scores of 75, 82, 95, 91, and 94 on his first five tests. Only the final exam remains, and the maximum of points that can be earned is 100. Is it possible for Josh to end the course with an A? A simple linear equation will give Josh his answer. Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce x widgets per month plus monthly operating charges; a car rental company charges a daily fee plus
an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems. Setting up a Linear Equation to Solve a Real-World Application To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write 0.10x. This expression represents a variable cost because it changes according to the number of miles driven. If a quantity is independent of a variable, we usually just add or subtract it, according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost C. When dealing with real-world applications, there are certain expressions that we can translate directly into math. Table 1 lists some common verbal expressions and their equivalent mathematical expressions. C = 0.10x + 50 Verbal One number exceeds another by a Twice a number One number is a more than another number One number is a less than twice another number The product of a number and a, decreased by b The quotient of a number and the number plus a is three times the number The product of three times a number and the number decreased by b is c Table 1 Translation to Math Operations x, x + a 2x x, x + a x, 2x βˆ’ a ax βˆ’ b x _____ x + a = 3x 3x(x βˆ’ b) = c Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.3 MODELS AND APPLICATIONS 103 How To… Given a real-world problem, model a linear equation to fit it. 1. Identify known quantities. 2. Assign a variable to represent the unknown quantity. 3. If there is more than one unknown quantity, find a way to write the second unknown in terms of the first. 4. Write an equation interpreting the words as mathematical operations. 5.
Solve the equation. Be sure the solution can be explained in words, including the units of measure. Example 1 Modeling a Linear Equation to Solve an Unknown Number Problem Find a linear equation to solve for the following unknown quantities: One number exceeds another number by 17 and their sum is 31. Find the two numbers. Solution Let x equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as x + 17. The sum of the two numbers is 31. We usually interpret the word is as an equal sign. x + (x + 17) = 31 2x + 17 = 31 2x = 14 x = 7 x + 17 = 7 + 17 = 24 Simplify and solve. The two numbers are 7 and 24. Try It #1 Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is 36, find the numbers. Example 2 Setting Up a Linear Equation to Solve a Real-World Application There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $.05/min talk-time. Company B charges a monthly service fee of $40 plus $.04/min talk-time. a. Write a linear equation that models the packages offered by both companies. b. If the average number of minutes used each month is 1,160, which company offers the better plan? c. If the average number of minutes used each month is 420, which company offers the better plan? d. How many minutes of talk-time would yield equal monthly statements from both companies? Solution a. The model for Company A can be written as A = 0.05x + 34. This includes the variable cost of 0.05x plus the monthly service charge of $34. Company B’s package charges a higher monthly fee of $40, but a lower variable cost of 0.04x. Company B’s model can be written as B = 0.04x + $40. b. If the average number of minutes used each month is 1,160, we have the following: Company A = 0.05(1,160) + 34 = 58 + 34 = 92 Company B = 0.04(1,160) + 40 = 46.4 + 40 = 86.4 So, Company B offers the lower monthly cost of $86.40 as compared with
the $92 monthly cost offered by Company A when the average number of minutes used each month is 1,160. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 104 CHAPTER 2 EQUATIONS AND INEQUALITIES c. If the average number of minutes used each month is 420, we have the following: Company A = 0.05(420) + 34 = 21 + 34 = 55 Company B = 0.04(420) + 40 = 16.8 + 40 = 56.8 If the average number of minutes used each month is 420, then Company A offers a lower monthly cost of $55 compared to Company B’s monthly cost of $56.80. d. To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of (x, y) coordinates: At what point are both the x-value and the y-value equal? We can find this point by setting the equations equal to each other and solving for x. 0.05x + 34 = 0.04x + 40 0.01x = 6 x = 600 Check the x-value in each equation. 0.05(600) + 34 = 64 0.04(600) + 40 = 64 Therefore, a monthly average of 600 talk-time minutes renders the plans equal. See Figure 2. y 90 80 70 60 50 40 30 0 B = 0.04x + 40 A = 0.05x + 34 100 200 300 400 500 600 700 800 900 1000 1100 1200 x Figure 2 Try It #2 Find a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company’s monthly expenses? Using a Formula to Solve a Real-World Application Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem’s question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the area of a rectangular region, A = LW; the perimeter of a rectangle, P = 2L + 2W
; and the volume of a rectangular solid, V = LWH. When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time. Example 3 Solving an Application Using a Formula It takes Andrew 30 min to drive to work in the morning. He drives home using the same route, but it takes 10 min longer, and he averages 10 mi/h less than in the morning. How far does Andrew drive to work? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.3 MODELS AND APPLICATIONS 105 Solution This is a distance problem, so we can use the formula d = rt, where distance equals rate multiplied by time. Note that when rate is given in mi/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution. 1 _ First, we identify the known and unknown quantities. Andrew’s morning drive to work takes 30 min, or h at 2 2 _ h, and his speed averages 10 mi/h less than the morning drive. Both trips rate r. His drive home takes 40 min, or 3 cover distance d. A table, such as Table 2, is often helpful for keeping track of information in these types of problems. Write two equations, one for each trip. To Work To Home d d d r r r βˆ’ 10 Table r βˆ’ 10) ( _ 3 To work To home As both equations equal the same distance, we set them equal to each other and solve for r. 2 1 ) ) = (r βˆ’ 10) ( r ( _ _ 2 3 20_ 20_ 3 20_ 3 20 _ 3 r = βˆ’ (βˆ’6) r = 40 We have solved for the rate of speed to work, 40 mph. Substituting 40 into the rate on the return trip yields 30 mi/h. Now we can answer the question. Substitute the rate back into either equation and solve for d. The distance between home and work is 20 mi. 1 ) d = 40 ( _ 2 = 20 Analysis Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for r. 2 1 ) ) = (r βˆ’ 10r βˆ’ 10 3r = 4(r βˆ’ 10) 3r = 4r βˆ’ 40 βˆ’r = βˆ’40 r = 40 Try It #3 On Saturday morning
, it took Jennifer 3.6 h to drive to her mother’s house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 h to return home. Her speed was 5 mi/h slower on Sunday than on Saturday. What was her speed on Sunday? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 106 CHAPTER 2 EQUATIONS AND INEQUALITIES Example 4 Solving a Perimeter Problem The perimeter of a rectangular outdoor patio is 54 ft. The length is 3 ft greater than the width. What are the dimensions of the patio? Solution The perimeter formula is standard: P = 2L + 2W. We have two unknown quantities, length and width. However, we can write the length in terms of the width as L = W + 3. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as in Figure 3. W L = W + 3 Figure 3 Now we can solve for the width and then calculate the length. P = 2L + 2W 54 = 2(W + 3) + 2W 54 = 2W + 6 + 2W 54 = 4W + 6 48 = 4W 12 = W (12 + 3) = L 15 = L The dimensions are L = 15 ft and W = 12 ft. Try It #4 Find the dimensions of a rectangle given that the perimeter is 110 cm and the length is 1 cm more than twice the width. Example 5 Solving an Area Problem The perimeter of a tablet of graph paper is 48 in.2. The length is 6 in. more than the width. Find the area of the graph paper. Solution The standard formula for area is A = LW; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one. We know that the length is 6 in. more than the width, so we can write length as L = W + 6. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length. P = 2L + 2W 48 = 2(W + 6) + 2W 48 = 2
W + 12 + 2W 48 = 4W + 12 36 = 4W 9 = W (9 + 6) = L 15 = L Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.3 MODELS AND APPLICATIONS 107 Now, we find the area given the dimensions of L = 15 in. and W = 9 in. A = LW A = 15(9) = 135 in.2 The area is 135 in.2. Try It #5 A game room has a perimeter of 70 ft. The length is five more than twice the width. How many ft2 of new carpeting should be ordered? Example 6 Solving a Volume Problem Find the dimensions of a shipping box given that the length is twice the width, the height is 8 inches, and the volume is 1,600 in 3. Solution The formula for the volume of a box is given as V = LWH, the product of length, width, and height. We are given that L = 2W, and H = 8. The volume is 1,600 cubic inches. V = LWH 1,600 = (2W)W(8) 1,600 = 16W2 100 = W2 10 = W The dimensions are L = 20 in., W = 10 in., and H = 8 in. Analysis describing width, we can use only the positive result. Note that the square root of W 2 would result in a positive and a negative value. However, because we are Access these online resources for additional instruction and practice with models and applications of linear equations. β€’ Problem Solving Using Linear Equations (http://openstaxcollege.org/l/lineqprobsolve) β€’ Problem Solving Using Equations (http://openstaxcollege.org/l/equationprsolve) β€’ Finding the Dimensions and Area Given the Perimeter (http://openstaxcollege.org/l/permareasolve) β€’ Find the Distance Between the Cities Using the distance = rate * time formula (http://openstaxcollege.org/l/ratetimesolve) β€’ Linear Equation Application (Write a cost equation) (http://openstaxcollege.org/l/lineqappl) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 108 CHAPTER 2 EQUATIONS AND INEQUALITIES 2.3 SECTION EXERC
ISES VERBAL 1. To set up a model linear equation to fit real-world applications, what should always be the first step? 2. Use your own words to describe this equation where n is a number: 5(n + 3) = 2n 3. If the total amount of money you had to invest was 4. If a man sawed a 10-ft board into two sections and $2,000 and you deposit x amount in one investment, how can you represent the remaining amount? one section was n ft long, how long would the other section be in terms of n? 5. If Bill was traveling v mi/h, how would you represent Daemon’s speed if he was traveling 10 mi/h faster? REAL-WORLD APPLICATIONS For the following exercises, use the information to find a linear algebraic equation model to use to answer the question being asked. 7. Beth and Ann are joking that their combined ages equal Sam’s age. If Beth is twice Ann’s age and Sam is 69 yr old, what are Beth and Ann’s ages? 6. Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113? 8. Ben originally filled out 8 more applications than Henry. Then each boy filled out 3 additional applications, bringing the total to 28. How many applications did each boy originally fill out? For the following exercises, use this scenario: Two different telephone carriers offer the following plans that a person is considering. Company A has a monthly fee of $20 and charges of $.05/min for calls. Company B has a monthly fee of $5 and charges $.10/min for calls. 9. Find the model of the total cost of Company A’s plan, 10. Find the model of the total cost of Company B’s plan, using m for the minutes. using m for the minutes. 11. Find out how many minutes of calling would make the two plans equal. 12. If the person makes a monthly average of 200 min of calls, which plan should for the person choose? For the following exercises, use this scenario: A wireless carrier offers the following plans that a person is considering. The Family Plan: $90 monthly fee, unlimited talk and text on up to 5 lines, and data charges of
$40 for each device for up to 2 GB of data per device. The Mobile Share Plan: $120 monthly fee for up to 10 devices, unlimited talk and text for all the lines, and data charges of $35 for each device up to a shared total of 10 GB of data. Use P for the number of devices that need data plans as part of their cost. 13. Find the model of the total cost of the Family Plan. 14. Find the model of the total cost of the Mobile Share Plan. 15. Assuming they stay under their data limit, find the number of devices that would make the two plans equal in cost. 16. If a family has 3 smart phones, which plan should they choose? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.3 SECTION EXERCISES 109 For exercises 17 and 18, use this scenario: A retired woman has $50,000 to invest but needs to make $6,000 a year from the interest to meet certain living expenses. One bond investment pays 15% annual interest. The rest of it she wants to put in a CD that pays 7%. 17. If we let x be the amount the woman invests in the 15% bond, how much will she be able to invest in the CD? 18. Set up and solve the equation for how much the woman should invest in each option to sustain a $6,000 annual return. 19. Two planes fly in opposite directions. One travels 450 mi/h and the other 550 mi/h. How long will it take before they are 4,000 mi apart? 20. Ben starts walking along a path at 4 mi/h. One and a half hours after Ben leaves, his sister Amanda begins jogging along the same path at 6 mi/h. How long will it be before Amanda catches up to Ben? 21. Fiora starts riding her bike at 20 mi/h. After a while, she slows down to 12 mi/h, and maintains that speed for the rest of the trip. The whole trip of 70 mi takes her 4.5 h. For what distance did she travel at 20 mi/h? 22. A chemistry teacher needs to mix a 30% salt solution with a 70% salt solution to make 20 qt of a 40% salt solution. How many quarts of each solution should the teacher mix to get the desired result? 23. Paul has $20,000 to invest.
His intent is to earn 11% interest on his investment. He can invest part of his money at 8% interest and part at 12% interest. How much does Paul need to invest in each option to make get a total 11% return on his $20,000? For the following exercises, use this scenario: A truck rental agency offers two kinds of plans. Plan A charges $75/wk plus $.10/mi driven. Plan B charges $100/wk plus $.05/mi driven. 24. Write the model equation for the cost of renting a 25. Write the model equation for the cost of renting a truck with plan A. truck with plan B. 26. Find the number of miles that would generate the 27. If Tim knows he has to travel 300 mi, which plan same cost for both plans. should he choose? For the following exercises, find the slope of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular. 28. A = P(1 + rt) is used to find the principal amount P deposited, earning r% interest, for t years. Use this to find what principal amount P David invested at a 3% rate for 20 yr if A = $8,000. 30. F = ma indicates that force (F) equals mass (m) times acceleration (a). Find the acceleration of a mass of 50 kg if a force of 12 N is exerted on it. mv2 _ R 29. The formula F = relates force (F), velocity (v), mass (m), and resistance (R). Find R when m = 45, v = 7, and F = 245. 31. Sum = is the formula for an infinite series 1 _ 1 βˆ’ r sum. If the sum is 5, find r. For the following exercises, solve for the given variable in the formula. After obtaining a new version of the formula, you will use it to solve a question. 32. Solve for W: P = 2L + 2W 1 1_ 1 _ q = _ p + 34. Solve for f: f 36. Solve for m in the slope-intercept formula: y = mx + b 33. Use the formula from the previous question to find the width, W, of a rectangle whose length is 15 and whose perimeter is 58. 35. Use the formula from the previous question to find f when p = 8 and q = 13. 37. Use the formula from the
previous question to find m when the coordinates of the point are (4, 7) and b = 12. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 110 CHAPTER 2 EQUATIONS AND INEQUALITIES 1 _ h(b1 + b2). 38. The area of a trapezoid is given by A = 2 Use the formula to find the area of a trapezoid with h = 6, b1 = 14, and b2 = 8. 1 _ h(b1 + b2) 39. Solve for h: A = 2 40. Use the formula from the previous question to find the height of a trapezoid with A = 150, b1 = 19, and b2 = 11. 41. Find the dimensions of an American football field. The length is 200 ft more than the width, and the perimeter is 1,040 ft. Find the length and width. Use the perimeter formula P = 2L + 2W. 42. Distance equals rate times time, d = rt. Find the distance Tom travels if he is moving at a rate of 55 mi/h for 3.5 h. 43. Using the formula in the previous exercise, find the distance that Susan travels if she is moving at a rate of 60 mi/h for 6.75 h. 44. What is the total distance that two people travel in 3 h if one of them is riding a bike at 15 mi/h and the other is walking at 3 mi/h? 1 _ 45. If the area model for a triangle is A = bh, find the 2 area of a triangle with a height of 16 in. and a base of 11 in. 1 _ 46. Solve for h: A = bh 2 47. Use the formula from the previous question to find the height to the nearest tenth of a triangle with a base of 15 and an area of 215. 48. The volume formula for a cylinder is V = Ο€r2 h. 49. Solve for h: V = Ο€r2h Using the symbol Ο€ in your answer, find the volume of a cylinder with a radius, r, of 4 cm and a height of 14 cm. 50. Use the formula from the previous question to find the height of a cylinder with a radius of 8 and a volume of 16Ο€ 52. Use the formula from the previous question to find the radius of a cylinder with
a height of 36 and a volume of 324Ο€. 54. Solve the formula from the previous question for Ο€. Notice why Ο€ is sometimes defined as the ratio of the circumference to its diameter. 51. Solve for r: V = Ο€r2h 53. The formula for the circumference of a circle is C = 2Ο€r. Find the circumference of a circle with a diameter of 12 in. (diameter = 2r). Use the symbol Ο€ in your final answer. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.4 COMPLEX NUMBERS 111 LEARNING OBJECTIVES In this section you will: β€’ Add and subtract complex numbers. β€’ Multiply complex numbers. 2.4 COMPLEX NUMBERS Figure 1 Discovered by Benoit Mandelbrot around 1980, the Mandelbrot Set is one of the most recognizable fractal images. The image is built on the theory of self-similarity and the operation of iteration. Zooming in on a fractal image brings many surprises, particularly in the high level of repetition of detail that appears as magnification increases. The equation that generates this image turns out to be rather simple. In order to better understand it, we need to become familiar with a new set of numbers. Keep in mind that the study of mathematics continuously builds upon itself. Negative integers, for example, fill a void left by the set of positive integers. The set of rational numbers, in turn, fills a void left by the set of integers. The set of real numbers fills a void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. In this section, we will explore a set of numbers that fills voids in the set of real numbers and find out how to work within it. Expressing Square Roots of Negative Numbers as Multiples of i We know how to find the square root of any positive real number. In a similar way, we can find the square root of any negative number. The difference is that the root is not real. If the value in the radicand is negative, the root is said to be an imaginary number. The imaginary number i is defined as the square root of βˆ’1. So, using properties of radicals, β€” βˆ’1 = i √ i2 = ( √ β€” 2 βˆ’1 ) = βˆ’1 We can write the square root of any negative
number as a multiple of i. Consider the square root of βˆ’49. We use 7i and not βˆ’7i because the principal root of 49 is the positive root. √ β€” βˆ’49 = √ = √ = 7i β€” 49 Δ‹ (βˆ’1) 49 √ βˆ’1 β€” β€” A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written a + bi where a is the real part and b is the imaginary part. For example, 5 + 2i is a complex number. So, too, is 3 + 4i √ 3. β€” 5 + 2i Imaginary numbers differ from real numbers in that a squared imaginary number produces a negative real number. Recall that when a positive real number is squared, the result is a positive real number and when a negative real number is squared, the result is also a positive real number. Complex numbers consist of real and imaginary numbers. Real part Imaginary part Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 112 CHAPTER 2 EQUATIONS AND INEQUALITIES imaginary and complex numbers A complex number is a number of the form a + bi where β€’ a is the real part of the complex number. β€’ b is the imaginary part of the complex number. If b = 0, then a + bi is a real number. If a = 0 and b is not equal to 0, the complex number is called a pure imaginary number. An imaginary number is an even root of a negative number. β€” How To… Given an imaginary number, express it in the standard form of a complex number. 1. Write √ 2. Express √ 3. Write √ βˆ’a as √ βˆ’1 as i. a Δ‹ i in simplest form. a √ βˆ’1. β€” β€” β€” β€” Example 1 Expressing an Imaginary Number in Standard Form β€” βˆ’9 in standard form. Express √ Solution In standard form, this is 0 + 3i. Try It #1 Express √ β€” βˆ’24 in standard form. √ β€” βˆ’9 = √ = 3i β€” 9 √ β€” βˆ’1 Plotting a Complex Number on the Complex Plane We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number, we need to address the two components of the number. We use the complex plane, which
is a coordinate system in which the horizontal axis represents the real component and the vertical axis represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs (a, b), where a represents the coordinate for the horizontal axis and b represents the coordinate for the vertical axis. Let’s consider the number βˆ’2 + 3i. The real part of the complex number is βˆ’2 and the imaginary part is 3. We plot the ordered pair (βˆ’2, 3) to represent the complex number βˆ’2 + 3i, as shown in Figure 2. βˆ’2 + 3i –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 2 complex plane In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis, as shown in Figure 3. imaginary real Figure 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.4 COMPLEX NUMBERS 113 How To… Given a complex number, represent its components on the complex plane. 1. Determine the real part and the imaginary part of the complex number. 2. Move along the horizontal axis to show the real part of the number. 3. Move parallel to the vertical axis to show the imaginary part of the number. 4. Plot the point. Example 2 Plotting a Complex Number on the Complex Plane Plot the complex number 3 βˆ’ 4i on the complex plane. Solution The real part of the complex number is 3, and the imaginary part is βˆ’4. We plot the ordered pair (3, βˆ’4) as shown in Figure 4. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure 4 Try It #2 Plot the complex number βˆ’4 βˆ’ i on the complex plane. Adding and Subtracting Complex Numbers Just as with real numbers, we can perform arithmetic operations on complex numbers. To add or subtract complex numbers, we combine the real parts and then combine the imaginary parts. complex numbers: addition and subtraction Adding complex numbers: Subtracting complex numbers: (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) βˆ’ (c + di) = (a βˆ’ c) + (b βˆ’ d)i