text stringlengths 235 3.08k |
|---|
7 2y = 7 β 19 18) y = x + 4 4y = 3x β 20) y = β 2x + 8 6y = β 7x β 22) x β 2y = 4x + 2y = 18 β β β 13 23) x 5y = 7 2x + 7y = β 20 24) 3x 4y = 15 β 7x + y = 4 5 26) 6x + 4y = 16 8 20 2x + y = β 3 β 13 28) 7x + 5y = 4y = x β 16 β β 5x 5y = 2x + y = 7 β β β β β y = β 23 25) 2x β 8y = β x β β 6x + y = 20 3y = 3x 27) β β 2... |
. 4y = 8 3x β 5x + 4y = 8x = 8 β β 24 16 8 Notice opposites in front of y β²s. Add columns. Solve for x, divide by 8 x = 2) + 4y = 10 + 4y = 2 We have our x! β 24 Plug into either original equation, simplify β 24 Add 10 to both sides β + 10 5( β β + 10 4y = 4 β 14 Divide by 4 4 7 y = β 2 7 2, β 2 Our Solution β Now we h... |
an important point, some problems we will have to multiply both equations by a constant (on both sides) to get the opposites we want. 5 to get opposites, 15y and β β Example 179. 3x + 6y = 2x + 9y = 9 26 β β We can get opposites in front of x, ο¬nd LCM of 6 and 9, 18y The LCM is 18. We will multiply to get 18y and β 3(... |
x 8 19y = 57 β β 19y = 57 19 19 y = 3 2x 2x 13 5(3) = β β 13 15 = β + 15 + 15 β 2x 2 = 2 2 x = 1 (1, 3) World View Note: The famous mathematical text, The Nine Chapters on the Mathematical Art, which was printed around 179 AD in China describes a formula very similar to Gaussian elimination which is very similar to the... |
= 13 β 6x + 9y = 3 9y = 9 β 9x β 6x β β 10 14 6y = 6y = β β 5y = 28 β β x x + 4y = β 7) 4x 4x 9) β β 11) 2x y = 5 β 5x + 2y = 17 β 28 β 13) 10x + 6y = 24 6x + y = 4 β 15) 2x + 4y = 24 12y = 8 4x 17) β 10x β 7x + 4y = 8y = 4 β 8 β β 19) 5x + 10y = 20 6x 5y = 3 β β β 3y = 12 5y = 20 18 10 β β 7x 6x 21) β β 23) 9x 5x β β... |
24 18 β β 30) 7x + 10y = 13 β 4x + 9y = 22 9x 21 + 12y 32=0 3 3 42y = 7 2 β y = 0 12x 34) β x 6 β 1 2 β β 150 4.4 Systems of Equations - Three Variables Objective: Solve systems of equations with three variables using addition/elimination. Solving systems of equations with 3 variables is very similar to how we solve s... |
= 2 We now have x! Plug this into either (A) or (B) Solve, divide by 2 (2) + 2z = 4 We plug it into (A), solve this equation, subtract 2 2 β 2 β 2z = 2 Divide by 2 2 z = 1 We now have z! Plug this and x into any original equation 2 3(2) + 2y (1) = β 2y + 5 = 5 β 2y = 2 y = Solve, subtract 5 1 We use the ο¬rst, multiply... |
12x β 55 This is our (A) equation β 6x + 2y 8x β β z = 16 Now use the second two equations (a diο¬erent pair) β y + 3z = 23 The LCM of 6 and β 8 is 24. β 4(6x + 2y z) = ( β 24x + 8y β 4 = β 16)4 Multiply the ο¬rst equation by 4 64 β 8x 3( β β 24x y + 3z) = (23)3 Multiply the second equation by 3 3y + 9z = 69 β β 24x + 8... |
we interpret the result is identical. Example 184. 4y + 3z = 5x 10x + 8y β 4 β β β 6z = 8 We will eliminate x, start with ο¬rst two equations 154 15x β 12y + 9z = 12 β 4y + 3z = 5x 10x + 8y β 4 β 6z = 8 β β LCM of 5 and 10 is 10. β 2(5x 4y + 3z) = 4(2) Multiply the ο¬rst equation by 2 β 10x β 8y + 6z = β 8 β 10x 8y + 6z... |
4 1 β 2) 2x + 3y = z 3x = 8z 1 β 5y + 7z = β 4) x + y + z = 2 1 4y + 5z = 31 6x 5x + 2y + 2z = 13 β 3) 3x + y β 5) x + 6y + 3z = 4 2x + y + 2z = 3 2y + z = 0 3x β 7 β β β 2y + 3z = 6 β 2x x 3 z = 0 9 + 2z = 0 β 11) 2x + y 3z = 1 β β x 4y + z = 6 4x + 16y + 4z = 24 β 3z = 0 13) 2x + y β x 4y + z = 0 4x + 16y + 4z = 0 β... |
4 7y + 3z = 6 β β 3x + y 4x β 22) 3x + y y 8x 2y 5x β β β β β z = 10 6z = β 5z = 1 3 156 23) 3x + 3y 6x + 2y 2y 5x 2z = 13 5z = 13 5z = 1 β β β β β 4y + 2z = 1 25) 3x β 2x + 3y x + 10y 1 3z = β 8z = 7 β β 27) m + 6n + 3p = 8 3m + 4n = β 5m + 7n = 1 3 10 β 30) 29) 2z = 2w + 2x + 2y β w + x + y + z = 5 3w + 2x + 2y + 4z... |
used for the that value each item has. The third column is used for the total value which we calculate by multiplying the number by the value. For example, if we have 7 dimes, each with a value of 10 cents, the total value is 7 10 = 70 cents. The last row of the table is for totals. We only will use the third row (als... |
one quarterβ (not 25 cents). On the penny it says βone centβ (not 1 cent). The rest of the world (Euros, Yen, Pesos, etc) all write the value as a number so people who donβt speak the language can easily use the coins. Ticket sales also have a value. Often diο¬erent types of tickets sell for diο¬erent prices (values). Th... |
. Here we will have statements such as βThere are twice as many dimes as nicklesβ. While it is clear that we need to multiply one variable by 2, it may not be clear which variable gets multiplied by 2. Generally the equations are backwards from the English sentence. If there are twice as many dimes, than we multiply th... |
the interest earned. This is shown in the following example. Example 189. A woman invests S4000 in two accounts, one at 6% interest, the other at 9% interest for one year. At the end of the year she had earned S270 in interest. How much did she have invested in each account? Invest Rate Interest 0.06 0.09 x y Invest R... |
+ 8000x + 320 = 1230 13000x + 320 = 1230 320 β 13000x = 910 13000 13000 320 β Multiply to ο¬ll in interest column. Be sure to distribute 8000(x + 0.04) Total interest was 1230. Last column gives our equation Combine like terms Subtract 320 from both sides Divide both sides by 13000 x = 0.07 We have our x, 7% interest (... |
collected was S132.50. How many of each type of ticket was sold? 11) There were 203 tickets sold for a volleyball game. For activity-card holders, the price was S1.25 each and for noncard holders the price was S2 each. The total amount of money collected was S310. How many of each type of ticket was sold? 12) At a loc... |
of nickels. The total value of all the coins is S2.75. Find the number of each type of coin in the bank. 22) A total of 26 bills are in a cash box. Some of the bills are one dollar bills, and the rest are ο¬ve dollar bills. The total amount of cash in the box is S50. Find the number of each type of bill in the cash box... |
st fund, ο¬nd the two rates of interest. 32) Millicent earned S435 last year in interest. If S3000 was invested at a certain rate of return and S4500 was invested in a fund with a rate that was 2% lower, ο¬nd the two rates of interest. 33) A total of S8500 is invested, part of it at 6% and the rest at 3.5%. The total int... |
twice as many nickels as dimes and 3 more dimes than quarters. How many coins of each kind were there? 43) 30 coins having a value of S3.30 consists of nickels, dimes and quarters. If there are twice as many quarters as dimes, how many coins of each kind were there? 44) A bag contains nickels, dimes and quarters havin... |
part to get total. be sure to distribute on the last row: (70 + x)0.6 35 + 0.8x = 42 + 0.6x The last column is our equation by adding 0.6x Move variables to one side, subtract 0.6x 0.6x β β 35 + 0.2x = 42 35 35 β 0.2x = 7 0.2 0.2 β Subtract 35 from both sides Divide both sides by 0.2 x = 35 We have our x! 35 mL must b... |
the mixture table together and get an equation to solve. However, here we are interested in systems of equations, with two unknown values. The following example is one such problem. Example 193. A farmer has two types of milk, one that is 24% butterfat and another which is 18% butterfat. How much of each should he use... |
105 Using our mixture table, use c and n for variables We do know the ο¬nal amount (30) and price, include this in the table Multiply amount by part to get totals c + n = 30 4c + 2.5n = 105 First equation comes from the ο¬rst column Second equation comes from the total column c + n = 30 We will solve this problem with s... |
of antifreeze and 24.5L of water w = 24.5 We have our w Our Solution 171 4.6 Practice - Mixture Problems Solve. 1) A tank contains 8000 liters of a solution that is 40% acid. How much water should be added to make a solution that is 30% acid? 2) How much antifreeze should be added to 5 quarts of a 30% mixture of antif... |
S4.50 per kilogram? 13) How many pounds of lima beans that cost 90c per pound must be mixed with 16 lb of corn that cost 50c per pound to make a mixture of vegetables that costs 65c per pound? 14) How many liters of a blue dye that costs S1.60 per liter must be mixed with 18 L of anil that costs S2.50 per liter to mak... |
lb. of the mixture? 24) A grocer is mixing 40 cent per lb. coο¬ee with 60 cent per lb. coο¬ee to make a mixture worth 54c per lb. How much of each kind of coο¬ee should be used to make 70 lb. of the mixture? 25) A grocer wishes to mix sugar at 9 cents per pound with sugar at 6 cents per pound to make 60 pounds at 7 cents... |
beans must be mixed with 80 lb of coο¬ee that is 30% java beans to make a coο¬ee blend that is 32% java beans? 35) The manager of a specialty food store combined almonds that cost S4.50 per pound with walnuts that cost S2.50 per pound. How many pounds of each were used to make a 100 lb mixture that cost S3.24 per pound?... |
5.1 Polynomials - Exponent Properties Objective: Simplify expressions using the properties of exponents. Problems with expoenents can often be simpliο¬ed using a few basic exponent properties. Exponents represent repeated multiplication. We will use this fact to discover the important properties. World View Note: The w... |
This is investigated in the following example. Example 202. 3 a2 This means we have a2 three times a2 a2 Β· Β· a2 Add exponents a6 Our solution A quicker method to arrive at the solution would have been to just multiply the 3 = a6. This is known as the power of a power rule of expoexponents, (a2)3 = a2 nents. Β· Power of... |
such as 53 does not mean we multipy 5 by 3, 5 = 125. This is shown in the next 5 rather we multiply 5 three times, 5 example. Γ Γ Example 208. (4x2y5)3 Put the exponent of 3 on each factor, multiplying powers 43x6y15 Evaluate 43 64x6y15 Our Solution In the previous example we did not put the 3 on the 4 and multipy to ... |
3)3 Use power rule in denominator 3m8n12 m6n9 Use quotient rule 3m2n3 Our solution Example 213. 3ab2(2a4b2)3 6a5b7 3ab2(8a12b6) 6a5b7 24a13b8 6a5b7 2 2 2 Simplify inside parenthesis ο¬rst, using power rule in numerator Simplify numerator using product rule Simplify using the quotient rule 4a8b)2 Now that the parenthesis... |
y3 4xy 26) (u2v2 2u4)3 Β· 2v3 Β· 2u3v 2b4 Β· 3a3b4 Β· 28) 3vu5 uv2 30) 2ba7 ba2 32) 2a2b2a7 (ba4)2 Β· 34) yx2 (y4)2 Β· 2y4 36) n3(n4)2 2mn 38) (2y 3x2)2 x2 2x2y4 Β· 40) 2x4y5 Β· 2z10 x2y7 (xy2z2)4 2q3 p3r4 2p3 Β· (qrp3)2 4 42) 182 5.2 Polynomials - Negative Exponents Objective: Simplify expressions with negative exponents using... |
exponent is now positive. Also, it is important to note a negative exponent does not mean the expression is negative, only that we need the reciprocal of the base. Following are the rules of negative exponents Rules of Negative Exponets: aβm = 1 m 1 aβm = am a b βm = bm am Negative exponents can be combined in several... |
3x3yβ Β· 5y3 5 12xβ 6xβ 2yβ 5y3 Simplify numerator with product rule, adding exponents Quotient rule to subtract exponets, be careful with negatives! ( ( 2) 5) 5) = ( 2) + 5 = 3 3) = 8 Negative exponent needs to move down to denominator ( β 3 = ( β 5 2x3yβ 2x3 y8 Our Solution 185 Example 219. 3 (3ab3)β 2aβ 2abβ 4b0 In ... |
Exponents Simplify. Your answer should contain only positive expontents. 1) 2x4yβ 2 3) (a4bβ Β· 3)3 (2xy3)4 2a3bβ 2 Β· 5) (2x2y2)4xβ 4 Β· 2xβ3y2 3x0 9) 3xβ3y3 11) 4xyβ3 Β· xβ4y0 Β· 4yβ1 2) 2aβ 3 2bβ (2a0b4)4 Β· (2x3)0 4) 2x3y2 Β· 6) (m0n3 2mβ 3nβ 3)0 Β· 1nβ 3y3 2x4yβ3 Β· 3x3y2 10) 3yx3 12) 4y β2 Β· 3xβ2yβ4 (2mβ 1nβ 3)4 3 Β· 7) (... |
xy0)β1 2xy4 28) 2yx2 xβ2 Β· (2x0y4)β1 30) uβ3vβ4 2v(2uβ3v4)0 bβ1 32) (2a4b0)0 34) 2b4cβ2 Β· aβ2b4 2aβ3b2 Β· (2b3c2)β4 36) ( (2xβ3y0zβ1)3 2x3 Β· xβ3y2 2 )β 38) 2q4 m2p2q4 (2mβ4p2)3 Β· 39) (yxβ4z2)β1 x2y3zβ1 z3 Β· 40) 2mpnβ3 (m0nβ4p2)3 2n2p0 Β· 187 5.3 Polynomials - Scientiο¬c Notation Objective: Multiply and divide expressions ... |
and negative exponents mean in standard notation we have a small number (less than one). Keeping this in mind, we can easily make conversions between standard notation and scientiο¬c notation. Example 221. Convert 14, 200 to scientiο¬c notation Put decimal after ο¬rst nonzero number 1.42 Exponent is how many times decima... |
6 Γ 189 Example 227. 4)3 Use power rule to deal with numbers and 10β²s separately Γ 10β (1.8 1.83 = 5.832 Evaluate 1.83 4)3 = 10β 10β 12 Multiply exponents 12 Our Solution 5.832 (10β Γ Often when we multiply or divide in scientiο¬c notation the end result is not in scientiο¬c notation. We will then have to convert the fro... |
2) 0.000744 4) 1.09 6) 15000 8) 2.56 x 102 10) 5 x 104 12) 6 x 10β 5 Simplify. Write each answer in scientiο¬c notation. 13) (7 x 10β 1)(2 x 10β 3) 15) (5.26 x 10β 5)(3.16 x 10β 2) 17) (2.6 x 10β 2)(6 x 10β 2) 101 19) 4.9 Γ 10β3 2.7 Γ 21) 5.33 9.62 Γ Γ 23) (5.5 10β6 10β2 25) (7.8 10β 5)2 10β 2)5 Γ Γ 27) (8.03 104)β 4 Γ... |
ials are made up of terms. Terms are a product of numbers and/or variables. For example, 5x, 2y2, 5, ab3c, and x are all terms. Terms are connected to each other by addition or subtraction. Expressions are often named based on the number of terms in them. A monomial has one term, such as 3x2. A binomial has two terms, ... |
death from cancer. Generally when working with polynomials we do not know the value of the variable, so we will try and simplify instead. The simplest operation with polynomials is addition. When adding polynomials we are mearly combining like terms. Consider the following example Example 232. (4x3 β 2x + 8) + (3x3 9x... |
5 β 1 β β β 7) x2 + 9x + 23 when x = 3 β 8) β 9) x4 6x3 + 41x2 6x3 + x2 β β β 32x + 11 when x = 6 24 when x = 6 10) m4 + 8m3 + 14m2 + 13m + 5 when m = 6 β 11) (5p 5p4) (8p 8p4) β β 12) (7m2 + 5m3) β (6m3 5m2) 13) (3n2 + n3) β β (2n3 β 7n2) β 14) (x2 + 5x3) + (7x2 + 3x3) 15) (8n + n4) (3n β 16) (3v4 + 1) + (5 β 4n4) β ... |
+ 2x4 + 7x3) + (6x3 33) (n β 5n4 + 7) + (n2 β 34) (8x2 + 2x4 + 7x3) + (7x4 7n2 4n4) β 8x4 n) β 7n4 7x2) β β 7x3 + 2x2) β β 8 + 2x + 6x3) 35) (8r4 β 36) (4x3 + x β 37) (2n2 + 7n4 5r3 + 5r2) + (2r2 + 2r3 7r4 + 1) 7x2) + (x2 β 2) + (2 + 2n3 + 4n2 + 2n4) 38) (7b3 39) (8 β β 4b + 4b4) β b + 7b3) (8b3 β β (3b4 + 7b β β 8n3)... |
x, y, and z 8x5y10z4 Our Solution In the previous example it is important to remember that the z has an exponent of 1 when no exponent is written. Thus for our answer the z has an exponent of 1 + 3 = 4. Be very careful with exponents in polynomials. If we are adding or subtracting the exponnets will stay the same, but... |
2x β 8x3 (2x 7x(2x β β β 5)(4x2 5) + 3(2x 14x2 + 35x + 6x 34x2 + 41x 8x3 β 5) β β 20x2 β β β β 7x + 3) Distribute (2x 5) through parenthesis β 5) Distribute again through each parenthesis 15 Combine like terms 15 Our Solution This process of multiplying by distributing can easily be reversed to do an important procedur... |
6x 8x3 β β 5(4x2) 5( β 7x) β β β 20x2 + 35x 34x2 + 41x β β β β 5(3) Multiply out each term 15 Combine like terms 15 Our Solution The second step of the FOIL method is often not written, for example, consider the previous example, a student will often go from the problem (4x + 7y)(3x 2y) β and do the multiplication men... |
. All three methods are shown side by side in the example. Example 244. (2x β y)(4x 5y) β 4x(2x 8x2 Distribute 5y(2x 10xy y) β 4xy β β 14xy β 5y2 y) β 5y2 β β 8x2 β 2x(4x) + 2x( FOIL 5y) β 10xy β β y( y(4x) 4xy + 5y2 β 14xy + 5y2 8x2 β 8x2 β 5y) β Rows y 2x 5y 4x β Γ β 10xy + 5y2 4xy 14xy + 5y2 β 8x2 β 8x2 β When we ar... |
+ 1)(x 4) β 10) (r + 8)(4r + 8) 12) (7n β 6)(n + 7) 14) (6a + 4)(a 8) β 6)(4x 16) (5x β β 18) (2u + 3v)(8u 1) 7v) β 19) (x + 3y)(3x + 4y) 20) (8u + 6v)(5u 8v) 21) (7x + 5y)(8x + 3y) 22) (5a + 8b)(a β 3b) β 7)(6r2 r + 5) 24) (4x + 8)(4x2 + 3x + 5) 23) (r β 25) (6n β 4)(2n2 β β 27) (6x + 3y)(6x2 2n + 5) 7xy + 4y2) β 29)... |
useful to us as our study of algebra continues. The ο¬rst shortcut is often called a sum and a diο¬erence. A sum and a diο¬erence is easily recognized as the numbers and variables are exactly the same, but the sign in the middle is diο¬erent (one sum, one diο¬erence). To illustrate the shortcut consider the following examp... |
a + b)2 Squared is same as multiplying by itself (a + b)(a + b) Distribute (a + b) a(a + b) + b(a + b) Distribute again through ο¬nal parenthesis a2 + ab + ab + b2 Combine like terms ab + ab a2 + 2ab + b2 Our Solution This problem also helps us ο¬nd our shortcut for multiplying. The ο¬rst term in the answer is the square ... |
to memory. The more familiar we are with them, the easier factoring, or multiplying in reverse, will be. The ο¬nal example covers both types of problems (two perfect squares, one positive, one negative), be sure to notice the diο¬erence between the examples and how each formula is used Example 255. (4x 7)(4x + 7) 49 β 1... |
3x) 14) (3y β 16) (1 + 5n)2 18) (v + 4)2 20) (1 β 22) (7k 6n)2 β 7)2 5)2 24) (4x β 26) (3a + 3b)2 28) (4m n)2 β 30) (8x + 5y)2 32) (m 7)2 β 34) (8n + 7)(8n 36) (b + 4)(b β 38) (7x + 7)2 7) β 4) 40) (3a β 8)(3a + 8) 204 5.7 Polynomials - Divide Polynomials Objective: Divide polynomials using long division. Dividing pol... |
review the (general) steps that are used with whole numbers that we will also use with polynomials Example 258. 4 631 Divide front numbers: = 1 | 6 4 4 1 631 Multiply this number by divisor: 1 | 4 β 23 Change the sign of this number (make it subtract) and combine Bring down next number 4 = 4 Β· 15 Repeat, divide front ... |
4x + 7 Bring down the next term β 4x Repeat, divide front terms: β x = 4 β x β 4 | β 32x + 7 5x2 3x3 β β 3x3 + 12x2 7x2 32x β 7x2 + 28x β x β 4 | β 32x + 7 3x2 + 7x 4 β 5x2 3x3 β β 3x3 + 12x2 7x2 32x β 7x2 + 28x β 4x + 7 Multiply this term by divisor: 16 Change the signs and combine β + 4x 4(x β 4) = β 4x + 16 β β β 9... |
due to positive) 3x2 β 10x + 25 + 3 2x + 4 Our Solution In both of the previous example the dividends had the exponents on our variable counting down, no exponent skipped, third power, second power, ο¬rst power, zero power (remember x0 = 1 so there is no variable on zero power). This is very important in long division, ... |
this ο¬nal example illustrates, just as in regular long division, sometimes we have no remainder in a problem. World View Note: Paolo Ruο¬ni was an Italian Mathematician of the early 19th century. In 1809 he was the ο¬rst to describe a process called synthetic division which could also be used to divide polynomials. 209 ... |
+ 9m 9 β 24) 2x2 3 3m β 5x β 2x + 3 8 β 26) 3v2 3v 32 9 β β 28) 4n2 38 23n β 4n + 5 β 36) 2n3 + 21n2 + 25n 2n + 3 38) 8m3 57m2 + 42 β 8m + 7 40) 2x3 + 12x2 + 4x 2x + 6 37 β 38b2 + 29b 4b 7 β 60 β 42) 24b3 β 210 55 β 30) 8k3 β 31) x3 41 β 26x x + 4 β 33) 3n3 + 9n2 64n β n + 6 68 β 32) x3 β 34) k3 β 66k2 + 12k + 37 8 k ... |
3x + 8). β β β To do this we have to be able to ο¬rst identify what is the GCF of a polynomial. We will ο¬rst introduce this by looking at ο¬nding the GCF of several numbers. To ο¬nd a GCF of sevearal numbers we are looking for the largest number that can be divided by each of the numbers. This can often be done with quic... |
the parenthesis 3x + 4) Our Solution β Example 266. 3x3y2z + 5x4y3z5 β 4xy4 GCF is xy2, divide each term by this 213 3x3y2z xy2 = 3x2z, 5x4y3z5 xy2 = 5x3yz5, β 4xy4 xy2 = xy2(3x2z + 5x3yz5 β 4y2 This is what is left in parenthesis β 4y2) Our Solution World View Note: The ο¬rst recorded algorithm for ο¬nding the greatest... |
of each expression. 1) 9 + 8b2 3) 45x2 5) 56 β 25 β 35p 7) 7ab 35a2b β 3a2b + 6a3b2 9) β 5x2 11) β 13) 20x4 5x3 15x4 β β 30x + 30 β 15) 28m4 + 40m3 + 8 2) x 5 β 4) 1 + 2n2 80y 6) 50x β 8) 27x2y5 72x3y2 β 10) 8x3y2 + 4x3 12) β 32n9 + 32n6 + 40n5 14) 21p6 + 30p2 + 27 16) β 10x4 + 20x2 + 12x 18) 27y7 + 12y2x + 9y2 17) 30... |
, a GCF does not have to be a monomial, it could be a binomial. To see this, consider the following two example. Example 270. 3ax x(3a β β 7bx Both have x in common, factor it out 7b) Our Solution Now the same problem, but instead of x we have (2a + 5b). Example 271. 3a(2a + 5b) 7b(2a + 5b) Both have (2a + 5b) in commo... |
3) Split problem into two groups GCF on left is 5b, on the right is 2 (2a + 3) is the same! Factor out this GCF (2a + 3)(5b + 2) Our Solution The key for grouping to work is after the GCF is factored out of the left and right, the two binomials must match exactly. If there is any diο¬erence between the two we either ha... |
two binomials match. β Example 276. 12ab 12ab 2a(6b 14a β 14a Split the problem into two groups GCF on left is 2a, on right, no GCF, use (6b β 7) is the same! Factor out this GCF 6b + 7 6b + 7 7) β 1) Our Solution β β 1(6b 7)(2a 1 β 7) (6b β β β β β Example 277. 6x3 3x2(2x 5 6x3 15x2 + 2x β 15x2 + 2x 5) + 1(2x (2x β β... |
, one with subtraction. If it happens with addition, for 218 example the binomials are (a + b) and (b + a), we donβt have to do any extra work. This is because addition is the same in either order (5 + 3 = 3 + 5 = 8). Example 279. 7 + y 7 + y 1(7 + y) 21x 3xy β 3xy 21x β 3x(y + 7) β β β (y + 7)(1 3x) Our Solution β Spl... |
3 + 21b2 35b 49 β 7) 3x3 + 15x2 + 2x + 10 β 2) 35x3 10x2 β β 56x + 16 4) 14v3 + 10v2 7v 5 β β 6) 6x3 β 48x2 + 5x 40 β 8) 28p3 + 21p2 + 20p + 15 35 β 12) 42r3 β 9) 35x3 β 11) 7xy β 28x2 20x + 16 β 49x + 5y 13) 32xy + 40x2 + 12y + 15x 15) 16xy 17) 2xy β 56x + 2y β 8x2 + 7y3 7 β 28y2x 19) 40xy + 35x 21) 32uv β β 8y2 β 20u... |
F of the left two terms is x and the GCF of the second two terms is 4. The way we will factor trinomials is to make them into a polynomial with four terms and then factor by grouping. This is shown in the following example, the same problem worked backwards β Example 283. x2 + 6x x(x + 6) x2 + 2x β 4x β 4(x + 6) β β (x... |
(x (x β β β β x(x β 20, add to 10 and 2, split the middle term 20 Want to multiply to 20 10) β Factor by grouping β 8 β 10)(x + 2) Our Solution Often when factoring we have two variables. These problems solve just like problems with one variable, using the coeο¬cients to decide how to split the middle term Example 287. ... |
this is the case then we can use this shortcut. This is shown in the next few examples. β β Example 290. x2 β 7x β 18 Want to multiply to 18, add to 9 and 2, write the factors β 7 β β 9)(x + 2) Our Solution (x β 223 Example 291. m2 β mn β 30n2 Want to multiply to 30, add to β 1 β 6, write the factors, donβ²t forget sec... |
+ 15 11) n2 β 13) p2 + 15p + 54 15) n2 β 15n + 56 17) u2 8uv + 15v2 β 19) m2 + 2mn 8n2 β 11xy + 18y2 21) x2 β 23) x2 + xy β 25) x2 + 4xy 12y2 12y2 β 27) 5a2 + 60a + 100 29) 6a2 + 24a 192 β 31) 6x2 + 18xy + 12y2 33) 6x2 + 96xy + 378y2 2) x2 + x 4) x2 + x 72 30 β β 6) x2 + 13x + 40 8) b2 17b + 70 β 10) x2 + 3x 12) a2 6a... |
front of x2. This meant the number was 1 and we were multiplying to 1c or just c. Now we will have a number in front of x2 so we will be looking for numbers that multiply to ac and add to b. Other than this, the process will be the same. Β· Example 294. 3x2 + 11x + 6 Multiply to ac or (3)(6) = 18, add to 11 3x2 + 9x + ... |
. Factoring out the GCF ο¬rst also has the added bonus of making the numbers smaller so the ac method becomes easier. Example 298. 18x3 + 33x2 3x[6x2 + 11x 4x β β β β 2(2x + 5)] 30x GCF = 3x, factor this out ο¬rst 10] Multiply to ac or (6)( 10) = β 10] The numbers are 15 and Factor by grouping β 3x[6x2 + 15x 3x[3x(2x + 5... |
4 β β 10) 7x2 + 29x 30 β 26k + 24 12) 5k2 β 14) 3r2 + 16r + 21 16) 3u2 + 13uv 10v2 18) 7x2 2xy β β 20) 5u2 + 31uv β 5y2 28v2 β 22) 10a2 54a β 24) 21n2 + 45n 36 54 β β 26) 4r2 + r β 28) 6p2 + 11p 3 7 β 30) 4r2 + 3r 7 β 32) 4m2 + 6mn + 6n2 34) 4x2 β 36) 18u2 6xy + 30y2 3uv β 36v2 β 37) 12x2 + 62xy + 70y2 38) 16x2 + 60xy... |
36 Multiply to 36, add to 0 12, 4 9, 6 36, 2 18, 3 1 Β· Β· Β· Prime, cannot factor Our Solution Β· Β· 6 No combinations that multiply to 36 add to 0 229 It turns out that a sum of squares is always prime. Sum of Squares: a2 + b2 = Prime A great example where we see a sum of squares comes from factoring a diο¬erence of 4th p... |
sign from the middle 6 β 3 and β β 230 Example 306. 4x2 + 20xy + 25y2 Multiply to 100, add to 20 The numbers are 10 and 10, the same! Perfect square (2x + 5y)2 Use square roots from ο¬rst and last terms and sign from the middle World View Note: The ο¬rst known record of work with polynomials comes from the Chinese aroun... |
important point. When we ο¬ll in the trinomialβs ο¬rst and last terms we square the cube roots 5p and 2r. Often students forget to square the number in addition to the variable. Notice that when done correctly, both get cubed. 231 Often after factoring a sum or diο¬erence of cubes, students want to factor the second fact... |
x2 13) 18a2 4 27 36 β 50b2 β 2a + 1 15) a2 β 17) x2 + 6x + 9 6x + 9 19) x2 β 21) 25p2 10p + 1 β 23) 25a2 + 30ab + 9b2 25) 4a2 27) 8x2 20ab + 25b2 β 24xy + 18y2 β m3 64 29) 8 β 31) x3 β 33) 216 β 35) 125a3 u3 64 β 37) 64x3 + 27y3 39) 54x3 + 250y3 41) a4 43) 16 45) x4 81 z4 y4 β β β 47) m4 81b4 β 9 1 2) x2 β 4) x2 β 6) 4... |
β a2 + b2 = Prime a3 + b3 = (a + b)(a2 ab + b2) β b3 = (a a3 β β b)(a2 + ab + b2) 3 terms: ac method, watch for perfect square! a2 + 2ab + b2 = (a + b)2 Multiply to ac and add to b 4 terms: grouping β’ β’ β’ We will use the above strategy to factor each of the following examples. Here the emphasis will be on which strate... |
be comfortable and conο¬dent not just with using all the factoring methods, but decided on which method to use. This is why practice is very important! 235 6.6 Practice - Factoring Strategy Factor each completely. 45yh β 2) 2x2 β 11x + 15 4) 16x2 + 48xy + 36y2 6) 20uv 60u3 5xv + 15xu2 β 8) 2x3 + 5x2y + 3y2x β 1) 24az 1... |
Solve by Factoring Objective: Solve quadratic equation by factoring and using the zero product rule. When solving linear equations such as 2x 5 = 21 we can solve for the variable directly by adding 5 and dividing by 2 to get 13. However, when we have x2 (or a higher power of x) we cannot just isolate the variable as w... |
x β 5 = 0 or x x β + 5 + 5 3) = 0 The numbers are β 3 = 0 β + 3 + 3 3 β Set each factor equal to zero Solve each equation 5 and β 8 β x = 5 or x = 3 Our Solution Example 320. x2 β (x β 7x + 3x x2 4x 7)(x + 3) = 21 = β 21 = β + 9 + 9 12 = 0 β β β 4x β x2 9 Not equal to zero, multiply ο¬rst, use FOIL 9 Combine like terms ... |
this. Example 323. 4x2 = 8x 8x Set equal to zero by moving the terms to left 8x Be careful, on the right side, they are not like terms! β 4x2 4x(x 4x = 0 or x 4 4 Factor out the GCF of 4x β 8x = 0 β 2) = 0 One factor must be zero β 2 = 0 β + 2 + 2 Set each factor equal to zero Solve each equation x = 0 or 2 Our Soluti... |
with x3. He kept his method a secret until another mathematician, Cardan, talked him out of his secret and published the results. To this day the formula is known as Cardanβs Formula. A question often asked is if it is possible to get rid of the square on the variable by taking the square root of both sides. While it ... |
+ 7a β β 9 = β 3 + 6a 27) 35x2 + 120x = 45 β 23 = 6k 29) 4k2 + 18k β 7 β 32) x2 + 10x + 30 = 6 31) 9x2 β 46 + 7x = 7x + 8x2 + 3 33) 2m2 + 19m + 40 = 2m β 35) 40p2 + 183p 168 = p + 5p2 β 34) 5n2 + 41n + 40 = 2 β 80 = 3x 36) 24x2 + 11x β 241 Chapter 7 : Rational Expressions 7.1 Reduce Rational Expressions.................. |
to aid in the use of their base 20 system as a place holder! Rational expressions are easily evaluated by simply substituting the value for the variable and using order of operations. Example 327. x2 4 β x2 + 6x + 8 when x = 6 β Substitute β 5 in for each variable 6)2 ( β 6)2 + 6( ( β 36 36 + 6( 4 6) + 8 4 6) + 8 β β ... |
ated by multiplication). they are terms (separated by + or β 5 β 245 7.1 Practice - Reduce Rational Expressions Evaluate 1) 4v + 2 6 when v = 4 3) 5) x 3 β x2 4x + 3 when x = β b + 2 β b2 + 4b + 4 when b = 0 4 4) 2) b 3b 3 9 when b = β β a + 2 2 β a2 + 3a + 2 when a = 1 β 3 when n = 4 β 6 6) n2 β n n β State the exclud... |
) 56x 48 β 24x2 + 56x + 32 43) 2x2 3x2 10x + 8 7x + 4 β β 45) 7n2 32n + 16 16 β 4n β 47) n2 2n + 1 β 6n + 6 49) 7a2 6a2 26a 45 β 34a + 20 β β 44) 50b 80 β 50b + 20 46) 35v + 35 21v + 7 48) 56x 48 β 24x2 + 56x + 32 50) 4k3 9k3 2k2 2k β 18k2 + 9k β β 247 7.2 Rational Expressions - Multiply & Divide Objective: Multiply an... |
) and (x 4) β Multiply across 249 (x 3)(x β β 3(x + 5) 4) Our Solution Again we follow the same pattern with division with the extra ο¬rst step of multiplying by the reciprocal. Example 336. 12 8 Γ· x β 2x β 5x2 + 15x x2 + x 2 β Multiply by the reciprocal x2 + x 2 5x2 + 15x β Factor each numerator and denominator x2 x2 β... |
m(7m 7(7m 5) 5) β β 9) 7r 7r(r + 10) Γ· r (r 6 6)2 β β 11) 25n + 25 5 4 30n + 30 Β· 13) x 15) x2 β 10 35x + 21 Γ· 6x β x + 5 β 7 7 35x + 21 x + 5 x 7 β Β· 17) 8k 24k2 40k Γ· β 15k 1 β 25 19) (n 8) β 21) 4m + 36 m + 9 Β· 3x 6 12x 23) β β 80 6 β 10n Β· m 5 β 5m2 24(x + 3) 12 6n β 13n + 42 n2 β β 27) n 6n 7 12 Β· 29) 27a + 36 9a ... |
x2 k 7 β k 30) k2 β β 12 Β· 32) 9x3 + 54x2 x2 + 5x 28k 56k β β x2 + 5x β 10x2 14 14 Β· β 7 9n + 54 10n + 50 34) n β 2n n2 β 36) 7x2 β 49x2 + 7x β 35 Γ· 66x + 80 72 Γ· β 7x2 + 39x 2 49x + 7x β β 70 72 15 25) b + 2 40b2 β 24b(5b 3) β 26) 21v2 + 16v 3v + 4 β 16 37) 10b2 30b + 20 Β· 30b + 20 2b2 + 10b 39) 7r2 24 53r β β 7r + 2 ... |
that is also divisible by 6 24 Our Solution When ο¬nding the LCD of several monomials we ο¬rst ο¬nd the LCD of the coeο¬cients, then use all variables and attach the highest exponent on each variable. Example 339. Find the LCD of 4x2y5 and 6x4y3z6 First ο¬nd the LCD of coeο¬cients 4 and 6 12 is the LCD of 4 and 6 12 x4y5z6 ... |
can build up a fractionβs denominator by multipliplying the numerator and denoinator by any factors that are not already in the denominator. Example 342. 5a 3a2b =? 6a5b3 Idenο¬ty what factors we need to match denominators 2a3b2 3 Β· 2 = 6 and we need three more aβ²s and two more bβ²s 5a 3a2b 2a3b2 2a3b2 Multiply numerato... |
First: (x + 3) Second: (x Identify which factors are missing 6) Multiply fractions by missing factors β 5x 6)(x + 1) x + 3 x + 3 (x β and x 2 β (x + 1)(x + 3) x x 6 6 β β Multiply numerators 5x2 + 15x 6)(x + 1)(x + 3) and (x (x β x2 8x + 12 6)(x + 1)(x + 3) β β Our Solution World View Note: When the Egyptians began wo... |
) x2 16 β 27) x + 1 x2 36 β,, 3x 8x + 16 x2 β 2x + 3 x2 + 12x + 36 29) 4x x β x2 3x x β 5 β x2, 2 x, β x β 3 6 6x 5x + 1 3x β β, 10 x 4 β 5 3x + 1 x β β 12 3x 6x + 8 β,, 2x x2 + 4x + 3 x β x2 + x 2 β, 5 x2 + 3x 20 10 β x2 x2 x2 22) 24) 26) 28) 30) 256 7.4 Rational Expressions - Add & Subtract Objective: Add and subtrac... |
addition and subtraction, a pair of legs walking in the direction one reads for addition, and a pair of legs walking in the opposite direction for subtraction.. When we donβt have a common denominator we will have to ο¬nd the least common denominator (LCD) and build up each fraction so the denominators match. The follo... |
3a 8(2a + 1) Add numerators 6a2 + 3a + 12 8(2a + 1) Our Solution With subtraction remember to add the opposite. Example 352 7x + 12 β x2 Add the opposite (distribute negative) 4 β + β x2 β 4)(x x 1 β 7x + 12 3) (x β β Factor denominators to ο¬nd LCD LCD is (x 4)(x β β 3), build up each fraction β + β x2 β x 1 β 7x + 12... |
β β a2 9 β 2z + 3z 2z + 1 β x2 + 3x + 2 + 3x β 2x β 3 1 41) 1 β x2 + 5x + 6 2 x2 + 3x + 2 7 x2 + 13x + 42 3 4z2 β 1 38) 40) 43) x2 2x + 7 2x β 3 β β 3x 2 β x2 + 6x + 5 44) 261 4 a2 + 5a 6 β 2) 4) 6) 3 8) x2 x 2 β β a2 + 3a 6x x 8 β 2 β 7 β x2 6 β a2 + 5a x + 4 xy2 + 3 8 + x 3a2 + 5a + 1 x2y β 12 9a β 1 3 10) x + 5 12)... |
integers, then we will consider how the process can be expanded to include expressions with variables. The ο¬rst method uses order of operations to simplify the numerator and denominator ο¬rst, then divide the two resulting fractions by multiplying by the reciprocal. Example 353 12 8 12 β 5 6 + 3 6 5 12 4 3 3 4 1 4 5 16... |
x + 4) x + 4 β 5(x + 4) + 2(x + 4) x + 4 Reduce fractions 2(x + 4) 3 5(x + 4) + 2 β Distribute 2x 3 8 5x + 20 + 2 β β Combine like terms 2x 5 β β 5x + 22 Our Solution The more fractions we have in our problem, the more we repeat the same process. Example 356. ab3 + 1 3 2 ab2 β 4 a2b + ab ab 1 ab β Idenο¬ty LCD (highest ... |
2 6x + 9 6x β 6x + 9 + x2 + 6x β 9 9 β β 12x β 2x2 + 18 12x 2(x2 + 9) β β x2 6x 9 β Reduce fractions FOIL Combine like terms Factor out 2 in denominator Divide out common factor 2 Our Solution If there are negative exponents in an expression we will have to ο¬rst convert these negative exponents into fractions. Remember... |
3 x2 x + 1)2 + 1 (x 1)2 β Simplify each of the following fractional expressions. 31) xβ xβ 2 yβ β 1 + yβ 2 1 33) xβ 3y xβ 2 β β 3 xyβ yβ 2 2 xβ 35) 1 + 9 6xβ 9 β x2 β 32) 2 2y + xyβ 2 2 xβ xβ yβ β 2 3 4xβ 4 1 + xβ xβ 2 β xβ xβ 3 + yβ 1yβ β 1 + yβ 2 4 β 34) 36) 2 xβ 267 7.6 Rational Expressions - Proportions Objective:... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.