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7 2y = 7 − 19 18) y = x + 4 4y = 3x − 20) y = − 2x + 8 6y = − 7x − 22) x − 2y = 4x + 2y = 18 − − − 13 23) x 5y = 7 2x + 7y = − 20 24) 3x 4y = 15 − 7x + y = 4 5 26) 6x + 4y = 16 8 20 2x + y = − 3 − 13 28) 7x + 5y = 4y = x − 16 − − 5x 5y = 2x + y = 7 − − − − − y = − 23 25) 2x − 8y = − x − − 6x + y = 20 3y = 3x 27) − − 29) 3x + y = 9 2x + 8y = − − 18 − 16 30) 144 31) 2x + y = 2 3x + 7y = 14 33) x + 5y = 15 3x + 2y = 6 − 32) 2x + y = − 5x + 3y = − 7 21 34) 2x + 3y = 7x + y = 3 − 10 35) 37) 2x + 4y = 2 − y = − 16 − 6x + 6y = 3y = 16 − 8x − 12 − 39) 2x + 3y = 16 7x y = 20 − − − − 2x + 2y = 7y = 5x 22 19 − − − 8x + 2y = 6 − 2x + 3y = 11 − − 4y = x 14 − 6x + 8y = 12 − − − 36) 38) 40) 145 4.3 Systems of Equations - Addition/Elimination Objective: Solve systems of equations using the addition/elimination method. When solving systems we have found that graphing is very limited when solving equations. We then considered a second method known as substituion. This is probably the most used idea in solving systems in various areas of algebra. However, substitution can get ugly if we don’t have a lone variable. This leads us to our second method for solving systems of equations. This method is known as either Elimination or Addition. We will set up the process in the following examples, then deﬁne the ﬁve step process we can use to solve by elimination. Example 177
. 4y = 8 3x − 5x + 4y = 8x = 8 − − 24 16 8 Notice opposites in front of y ′s. Add columns. Solve for x, divide by 8 x = 2) + 4y = 10 + 4y = 2 We have our x! − 24 Plug into either original equation, simplify − 24 Add 10 to both sides − + 10 5( − − + 10 4y = 4 − 14 Divide by 4 4 7 y = − 2 7 2, − 2 Our Solution − Now we have our y! 4y and 4y. In the previous example one variable had opposites in front of it, Adding these together eliminated the y completely. This allowed us to solve for the x. This is the idea behind the addition method. However, generally we won’t have opposites in front of one of the variables. In this case we will manipulate the equations to get the opposites we want by multiplying one or both equations (on both sides!). This is shown in the next example. − Example 178. − 6x + 5y = 22 2x + 3y = 2 We can get opposites in front of x, by multiplying the second equation by 3, to get 6x and + 6x − 3(2x + 3y) = (2)3 Distribute to get new second equation. 146 − 6x + 9y = 6 New second equation 6x + 5y = 22 First equation still the same, add 14y = 28 Divide both sides by 14 14 14 y = 2 We have our y! 2x + 3(2) = 2 Plug into one of the original equations, simplify Subtract 6 from both sides 2x + 6 = 2 6 6 − − 2x = 4 Divide both sides by 2 − 2 2 x = 2 We also have our x! ( − 2, 2) Our Solution − − When we looked at the x terms, 6x and 2x we decided to multiply the 2x by 3 to get the opposites we were looking for. What we are looking for with our opposites is the least common multiple (LCM) of the coeﬃcients. We also could have solved the above problem by looking at the terms with y, 5y and 3y. The LCM of 3 and 5 is 15. So we would want to multiply both equations, the 5y by 3, and the 3y by 15y. This illustrates
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 − − Example 179. 3x + 6y = 2x + 9y = 9 26 − − We can get opposites in front of x, ﬁnd LCM of 6 and 9, 18y The LCM is 18. We will multiply to get 18y and − 3(3x + 6y) = ( 9x + 18y = 9)3 Multiply the ﬁrst equation by 3, both sides! 27 − − − 2(2x + 9y) = ( 4x − − − 2) Multiply the second equation by 26)( − 18y = 52 2, both sides! − 9x + 18y = 27 Add two new equations together − 18y = 52 − − 4x 5x 5 = 25 Divide both sides by 5 5 x = 5 We have our solution for x 3(5) + 6y = 15 + 6y = 15 9 Plug into either original equation, simplify − 9 − 15 − Subtract 15 from both sides − 147 6y = 6 − y = (5, − − 24 Divide both sides by 6 6 4 Now we have our solution for y 4) Our Solution It is important for each problem as we get started that all variables and constants are lined up before we start multiplying and adding equations. This is illustrated in the next example which includes the ﬁve steps we will go through to solve a problem using elimination. Problem 2x 5y = − − 3y + 4 = 13 5x − Second Equation: − 1. Line up the variables and constants 2. Multiply to get opposites (use LCD) 3. Add 4. Solve 5. Plug into either original and solve Solution 4 − 3y + 4 = − 3y = 5y = 3y = 5x − − 4 + 5x + 5x 5x 4 2x 13 5x 4 First Equation: multiply by 5(2x 5y) = ( 5) 10x + 25y = 65 − − − − − − 13)( − − − − − 5 − Second Equation: multiply by 2 2(5x 10x 3y) = ( − 6y = 8 4)2 − − − − 10x + 25y = 65 6y = 10
x 8 19y = 57 − − 19y = 57 19 19 y = 3 2x 2x 13 5(3) = − − 13 15 = − + 15 + 15 − 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 addition method. 148 Just as with graphing and substution, it is possible to have no solution or inﬁnite solutions with elimination. Just as with substitution, if the variables all disappear from our problem, a true statment will indicate inﬁnite solutions and a false statment will indicate no solution. Example 180. 2x 5y = 3 6x + 15y = − − To get opposites in front of x, multiply ﬁrst equation by 3 9 − 3(2x 5y) = (3)3 Distribute − 6x 15y = 9 − 6x 15y = 9 Add equations together − 6x + 15y = 9 − − 0 = 0 True statement Inﬁnite solutions Our Solution Example 181. 4x 6x − − 6y = 8 9y = 15 LCM for x′s is 12. 3(4x 6y) = (8)3 Multiply ﬁrst equation by 3 − 12x − 18y = 24 2(6x − 9y) = (15)( − 12x + 18y = − 2) Multiply second equation by 30 2 − − − 12x 18y = 24 Add both new equations together − − 12x + 18y = − 0 = − 30 6 False statement No Solution Our Solution We have covered three diﬀerent methods that can be used to solve a system of two equations with two variables. While all three can be used to solve any system, graphing works great for small integer solutions. Substitution works great when we have a lone variable, and addition works great when the other two methods fail. As each method has its own strengths, it is important you are familiar with all three methods. 149 4.3 Practice - Addition/Elimination Solve each system by elimination. 28 − 22 − 1) 4x + 2y = 0 9y = 4x − − 3) 5) 9x + 5y = 5y
= 13 − 6x + 9y = 3 9y = 9 − 9x − 6x − − 10 14 6y = 6y = − − 5y = 28 − − x x + 4y = − 7) 4x 4x 9) − − 11) 2x y = 5 − 5x + 2y = 17 − 28 − 13) 10x + 6y = 24 6x + y = 4 − 15) 2x + 4y = 24 12y = 8 4x 17) − 10x − 7x + 4y = 8y = 4 − 8 − − 19) 5x + 10y = 20 6x 5y = 3 − − − 3y = 12 5y = 20 18 10 − − 7x 6x 21) − − 23) 9x 5x − − 2y = 7y = − − 25) 9x + 6y = 21 9y = 28 − 10x − 7x + 5y = 3x 8 − 3y = 12 − − − − 8x 8y = − − 10x + 9y = 1 − 8 27) 29) 31) 9y = 7 x − 18y + 4x = − 33) 0 = 9x + 5y y = 2 7 x 26 − 2) 4) − − 7x + y = y = 9x x − 2y = − − x + 2y = 7 10 22 7 − − − 6) 5x 5x − − 8) 10) 12) 15 15 5y = 5y = − − − − 3x + 3y = 3x + 9y = 12 24 − − 5y = 0 10y = 10x 10x − − 5x + 6y = 2y = 5 − − − x 30 − 17 − − 14) x + 3y = 1 − − 10x + 6y = 10 6x + 4y = 12 − 12x + 6y = 18 16) 18) 20) 22) − − − 3x 6x + 4y = 4 y = 26 3x − 9x 5y = 19 − 7y = − 11 − − 5x + 4y = 4 10y = 7x 10 − − − 24) 3x + 7y = 4x + 6y = − 8 4 − − 5y = 12 4x 10x + 6y = 30 26) − − − 28) 8x + 7y = 6x + 3y =
24 18 − − 30) 7x + 10y = 13 − 4x + 9y = 22 9x 21 + 12y 32=0 3 3 42y = 7 2 − y = 0 12x 34) − x 6 − 1 2 − − 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 systems with two variables. When we had two variables we reduced the system down to one with only one variable (by substitution or addition). With three variables we will reduce the system down to one with two variables (usually by addition), which we can then solve by either addition or substitution. To reduce from three variables down to two it is very important to keep the work organized. We will use addition with two equations to eliminate one variable. This new equation we will call (A). Then we will use a diﬀerent pair of equations and use addition to eliminate the same variable. This second new equation we will call (B). Once we have done this we will have two equations (A) and (B) with the same two variables that we can solve using either method. This is shown in the following examples. Example 182. z = 1 − − 3x + 2y 2x − 5x + 2y − 2y + 3z = 5 We will eliminate y using two diﬀerent pairs of equations z = 3 − 151 z = 1 Using the ﬁrst two equations, − − 2y + 3z = 5 Add the ﬁrst two equations 3x + 2y 2x x − − (A) + 2z = 4 This is equation (A), our ﬁrst equation − (B) 2x − 5x + 2y 3x 2y + 3z = 5 Using the second two equations z = 3 Add the second two equations − + 2z = 8 This is equation (B), our second equation (A) x + 2z = 4 Using (A) and (B) we will solve this system. (B) 3x + 2z = 8 We will solve by addition − 1(x + 2z) = (4)( 2z = x 1) Multiply (A) by Add to the second equation, unchanged x 2z = − 3x + 2z = 8 2x = 4 2 2 x
= 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 − 2 − 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) = − 2y + 5 = 5 − 2y = 2 y = Solve, subtract 5 1 We use the ﬁrst, multiply 3(2) = 6 and combine with − 1 − 5 − 6 Divide by 2 − 2 3 We now have y! − 1 − (2, − 3, 1) Our Solution As we are solving for x, y, and z we will have an ordered triplet (x, y, z) instead of 152 just the ordered pair (x, y). In this above problem, y was easily eliminated using the addition method. However, sometimes we may have to do a bit of work to get a variable to eliminate. Just as with addition of two equations, we may have to multiply equations by something on both sides to get the opposites we want so a variable eliminates. As we do this remmeber it is improtant to eliminate the same variable both times using two diﬀerent pairs of equations. Example 183. 3y + 2z = z = 4x − 6x + 2y 8x − − − − 29 No variable will easily eliminate. 16 We could choose any variable, so we chose x − y + 3z = 23 We will eliminate x twice. 4x − 6x + 2y 3y + 2z = z = − 3(4x 3y + 2z) = ( − 12x − 9y + 6z = − 29 16 Make the ﬁrst equation have 12x, the second Start with ﬁrst two equations. LCM of 4 and 6 is 12. 12x − − − 29)3 Multiply the ﬁrst equation by 3 87 − 2(6x + 2y − z) = ( − 12x − − 16)( 4y + 2z = 32 − − 2) Multiply the second equation by 2 − 12x 9y + 6z = 87 Add these two equations together − 4y + 2z = 32 − (A) − 13y + 8z = −
12x − 55 This is our (A) equation − 6x + 2y 8x − − z = 16 Now use the second two equations (a diﬀerent pair) − y + 3z = 23 The LCM of 6 and − 8 is 24. − 4(6x + 2y z) = ( − 24x + 8y − 4 = − 16)4 Multiply the ﬁrst equation by 4 64 − 8x 3( − − 24x y + 3z) = (23)3 Multiply the second equation by 3 3y + 9z = 69 − − 24x + 8y 24x − − 64 Add these two equations together − − 4 = 3y + 9z = 69 5y + 5z = 5 This is our (B) equation (B) 153 (A) (B) − 13y + 8z = − 5y + 5z = 5 55 Using (A) and (B) we will solve this system The second equation is solved for z to use substitution − − − Solving for z, subtract 5y 5y + 5z = 5 5y 5y 5y Divide each term by 5 5z = 5 5 5 5 z = 1 y Plug into untouched equation − y) = 55 Distribute − − 8y = 55 Combine like terms − − 21y + 8 = 55 − 8 − 63 Divide by − 21 − y = 3 We have our y! Plug this into z = equations 8 − 21y = 21 Subtract 8 13y 8y 21 − − − − − − 13y + 8(1 13y + 8 − − z = 1 (3) Evaluate − z = − 2 We have z, now ﬁnd x from original equation. 4x − 3(3) + 2( 4x 29 Multiply and combine like terms 29 Add 13 2) = 13 = − − + 13 + 13 − − 4x = 4 − x = − 16 Divide by 4 4 4 We have our x! 4, 3, ( − − 2) Our Solution! World View Note: Around 250, The Nine Chapters on the Mathematical Art were published in China. This book had 246 problems, and chapter 8 was about solving systems of equations. One problem had four equations with ﬁve variables! Just as with two variables and two equations, we can have special cases come up with three variables and three equations. The way
we interpret the result is identical. Example 184. 4y + 3z = 5x 10x + 8y − 4 − − − 6z = 8 We will eliminate x, start with ﬁrst two equations 154 15x − 12y + 9z = 12 − 4y + 3z = 5x 10x + 8y − 4 − 6z = 8 − − LCM of 5 and 10 is 10. − 2(5x 4y + 3z) = 4(2) Multiply the ﬁrst equation by 2 − 10x − 8y + 6z = − 8 − 10x 8y + 6z = 8 Add this to the second equation, unchanged − 10x + 8y − − 6 = 8 0 = 0 A true statment − Inﬁnite Solutions Our Solution Example 185. 3x − 9x + 12y 4y + z = 2 3z = 4x − 2y − z = 3 − − 5 We will eliminate z, starting with the ﬁrst two equations 3x 4y + z = 2 The LCM of 1 and − 9x + 12y 3z = 5 − − 3 is 3 − − − 3(3x 4y + z) = (2)3 Multiply the ﬁrst equation by 3 − 9x − 12y + 3z = 6 12y + 3z = 6 Add this to the second equation, unchanged 9x 9x + 12y − − − 3z = 5 − 0 = 1 A false statement No Solution ∅ Our Solution Equations with three (or more) variables are not any more diﬃcult than two variables if we are careful to keep our information organized and eliminate the same variable twice using two diﬀerent pairs of equations. It is possible to solve each system several diﬀerent ways. We can use diﬀerent pairs of equations or eliminate variables in diﬀerent orders, but as long as our information is organized and our algebra is correct, we will arrive at the same ﬁnal solution. 155 4.4 Practice - Three Variables Solve each of the following systems of equation. 1) a 1 2b + c = 5 c = − 2c = − 2a + b − 3a + 3b − − z = 11 − x + 3y = z + 13 3z = 11 x + y
4 1 − 2) 2x + 3y = z 3x = 8z 1 − 5y + 7z = − 4) x + y + z = 2 1 4y + 5z = 31 6x 5x + 2y + 2z = 13 − 3) 3x + y − 5) x + 6y + 3z = 4 2x + y + 2z = 3 2y + z = 0 3x − 7 − − − 2y + 3z = 6 − 2x x 3 z = 0 9 + 2z = 0 − 11) 2x + y 3z = 1 − − x 4y + z = 6 4x + 16y + 4z = 24 − 3z = 0 13) 2x + y − x 4y + z = 0 4x + 16y + 4z = 0 − 15) 3x + 2y + 2z = 3 x + 2y 2x z = 5 4y + z = 0 − − 2y + 3z = 4 y + z = 1 17) x 2x − 4x + y + z = 1 − − y + 2z = 0 2y + 3z = 2y + z = 1 3 − − 19) x − x − 2x − 21) 4x − 5x + 9y 9x + 8y 3y + 2z = 40 7z = 47 3z = 97 − − 6) x − y + 2z = 3 − x + 2y + 3z = 4 2x + y + z = 3 − 8) x + y z = 0 − x + 2y − 2x + y + z = 0 4z = 0 10) x + 2y 4x 5x z = 4 3y + z = 8 y = 12 − − − 12) 4x + 12y + 16z = 4 3x + 4y + 5z = 3 x + 8y + 11z = 1 14) 4x + 12y + 16z = 0 3x + 4y + 5z = 0 x + 8y + 11z = 0 16) p + q + r = 1 p + 2q + 3r = 4 4p + 5q + 6r = 7 18) x + 2y 2x 3x − − − − 3z = 9 y + 2z = y − 4z = 3 8 20) 4x 7y + 3z = 1 2z =
4 7y + 3z = 6 − − 3x + y 4x − 22) 3x + y y 8x 2y 5x − − − − − z = 10 6z = − 5z = 1 3 156 23) 3x + 3y 6x + 2y 2y 5x 2z = 13 5z = 13 5z = 1 − − − − − 4y + 2z = 1 25) 3x − 2x + 3y x + 10y 1 3z = − 8z = 7 − − 27) m + 6n + 3p = 8 3m + 4n = − 5m + 7n = 1 3 10 − 30) 29) 2z = 2w + 2x + 2y − w + x + y + z = 5 3w + 2x + 2y + 4z = − − w + 3x − − 2y + 2z = 11 − 6 31 + 2x + 2y + 4z = 1 w + x y w + 3x + y − z = z = 6 2 − − − − − − 24) 2x 3y + 5z = 1 − 3x + 2y 4x + 7y − − 26) 2x + y = z z = 4 7z = 7 4x + z = 4y y = x + 1 28) 3x + 2y = z + 2 y = 1 3z = 2x 2y − − w + 2x − w + x + y 3y + = 22 14 − − 32 + 2x + 2y + z = 5 w + 3x + y 4 2w + x + y z = − 3z = − − − − − 7 − 157 4.5 Systems of Equations - Value Problems Objective: Solve value problems by setting up a system of equations. One application of system of equations are known as value problems. Value problems are ones where each variable has a value attached to it. For example, if our variable is the number of nickles in a person’s pocket, those nickles would have a value of ﬁve cents each. We will use a table to help us set up and solve value problems. The basic structure of the table is shown below. Number Value Total Item 1 Item 2 Total The ﬁrst column in the table is used for the number of things we have. Quite often, this will be our variables. The second column is
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 (also marked total) for the totals that · 158 are given to use. This means sometimes this row may have some blanks in it. Once the table is ﬁlled in we can easily make equations by adding each column, setting it equal to the total at the bottom of the column. This is shown in the following example. Example 186. In a child’s bank are 11 coins that have a value of S1.85. The coins are either quarters or dimes. How many coins each does child have? Quarter Dime Total Quarter Dime Total Quarter Dime Total Number Value Total q d 25 10 Using value table, use q for quarters, d for dimes Each quarter′s value is 25 cents, dime′s is 10 cents Number Value Total 25q 10d 25 10 q d Number Value Total 25q 10d 185 q d 11 25 10 Multiply number by value to get totals We have 11 coins total. This is the number total. We have 1.85 for the ﬁnal total, Write ﬁnal total in cents (185) Because 25 and 10 are cents q + d = 11 25q + 10d = 185 First and last columns are our equations by adding Solve by either addition or substitution. − 10(q + d) = (11)( 10d = 10q − − Using addition, multiply ﬁrst equation by 10 − 10) 110 − − − 10q − − 10d = 110 25q + 10d = 185 15q = 75 15 15 Add together equations Divide both sides by 15 q = 5 We have our q, number of quarters is 5 (5) + d = 11 5 5 Plug into one of original equations Subtract 5 from both sides d = 6 We have our d, number of dimes is 6 − − 159 5 quarters and 6 dimes Our Solution World View Note: American coins are the only coins that do not state the value of the coin. On the back of the dime it says “one dime” (not 10 cents). On the back of the quarter it says “
one quarter” (not 25 cents). On the penny it says “one cent” (not 1 cent). The rest of the world (Euros, Yen, Pesos, etc) all write the value as a number so people who don’t speak the language can easily use the coins. Ticket sales also have a value. Often diﬀerent types of tickets sell for diﬀerent prices (values). These problems can be solve in much the same way. Example 187. There were 41 tickets sold for an event. Tickets for children cost S1.50 and tickets for adults cost S2.00. Total receipts for the event were S73.50. How many of each type of ticket were sold? Child Adult Total Child Adult Total Child Adult Total Number Value Total c a 1.5 2 Using our value table, c for child, a for adult Child tickets have value 1.50, adult value is 2.00 (we can drop the zeros after the decimal point) Number Value Total 1.5c 2a 1.5 2 c a Number Value Total 1.5c 2a 73.5 c a 41 1.5 2 Multiply number by value to get totals We have 41 tickets sold. This is our number total The ﬁnal total was 73.50 Write in dollars as 1.5 and 2 are also dollars c + a = 41 First and last columns are our equations by adding 1.5c + 2a = 73.5 We can solve by either addition or substitution − − − Solve for a by subtracting c c + a = 41 We will solve by substitution. c c a = 41 c c) = 73.5 2c = 73.5 0.5c + 82 = 73.5 82 − 0.5c = Substitute into untouched equation Distribute Combine like terms Subtract 82 from both sides Divide both sides by 82 8.5 − − 0.5 − − − − 1.5c + 2(41 1.5c + 82 − 160 0.5 − a = 41 0.5 − c = 17 We have c, number of child tickets is 17 (17) Plug into a = equation to ﬁnd a − a = 24 We have our a, number of adult tickets is 24 17 child tickets and 24 adult tickets Our Solution Some problems will not give us the total number of items we have. Instead they will give a relationship between the items
. Here we will have statements such as “There are twice as many dimes as nickles”. 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 the other variable (nickels) by two. So the equation would be d = 2n. This type of problem is in the next example. Example 188. A man has a collection of stamps made up of 5 cent stamps and 8 cent stamps. There are three times as many 8 cent stamps as 5 cent stamps. The total value of all the stamps is S3.48. How many of each stamp does he have? Number Value Total f 3f 5 8 5f 24f 348 Number Value Total f e 5 8 5f 8e Number Value Total f e 5 8 5f 8e 348 Five Eight Total Five Eight Total Five Eight Total Use value table, f for ﬁve cent stamp, and e for eight Also list value of each stamp under value column Multiply number by value to get total The ﬁnal total was 338(written in cents) We do not know the total number, this is left blank. e = 3f 5f + 8e = 348 3 times as many 8 cent stamples as 5 cent stamps Total column gives second equation 5f + 8(3f ) = 348 Substitution, substitute ﬁrst equation in second 5f + 24f = 348 Multiply ﬁrst 29f = 348 29 29 Combine like terms Divide both sides by 39 f = 12 We have f. There are 12 ﬁve cent stamps e = 3(12) Plug into ﬁrst equation 161 12 ﬁve cent, 36 eight cent stamps Our Solution e = 36 We have e, There are 36 eight cent stamps The same process for solving value problems can be applied to solving interest problems. Our table titles will be adjusted slightly as we do so. Invest Rate Interest Account 1 Account 2 Total Our ﬁrst column is for the amount invested in each account. The second column is the interest rate earned (written as a decimal - move decimal point twice left), and the last column is for the amount of interset earned. Just as before, we multiply the investment amount by the rate to ﬁnd the ﬁnal column,
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 Rate Interest 0.06x 0.06 0.09y 0.09 x y Account 1 Account 2 Total Account 1 Account 2 Total Account 1 Account 2 Total Invest Rate Interest 0.06x 0.06 0.09y 0.09 270 x y 4000 Use our investment table, x and y for accounts Fill in interest rates as decimals Multiply across to ﬁnd interest earned. Total investment is 4000, Total interest was 276 x + y = 4000 0.06x + 0.09y = 270 First and last column give our two equations Solve by either substitution or addition − 0.06(x + y) = (4000)( 0.06x − 0.06y = − − − 0.06) Use Addition, multiply ﬁrst equation by 0.06 − 240 162 − 0.06x − − Add equations together 0.06y = 240 0.06x + 0.09y = 270 0.03y = 30 0.03 0.03 y = 1000 We have y, S1000 invested at 9% Divide both sides by 0.03 x + 1000 = 4000 1000 1000 − − Plug into original equation Subtract 1000 from both sides x = 3000 We have x, S3000 invested at 6% S1000 at 9% and S3000 at 6% Our Solution The same process can be used to ﬁnd an unknown interest rate. Example 190. John invests S5000 in one account and S8000 in an account paying 4% more in interest. He earned S1230 in interest after one year. At what rates did he invest? Interest Invest Rate 5000 8000 x + 0.04 x Account 1 Account 2 Total Our investment table. Use x for ﬁrst rate The second rate is 4% higher, or x + 0.04 Be sure to write this rate as a decimal! Account 1 Account 2 Total Account 2 Account 2 Total Invest Rate 5000 8000 x + 0.04 8000x + 320 Interest 5000x x Invest Rate 5000 8000 x + 0.04 8000x + 320 Interest 5000x x 1230 5000x
+ 8000x + 320 = 1230 13000x + 320 = 1230 320 − 13000x = 910 13000 13000 320 − Multiply to ﬁll 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 (0.07) + 0.04 0.11 S5000 at 7% and S8000 at 11% Second account is 4% higher The account with S8000 is at 11% Our Solution 163 4.5 Practice - Value Problems Solve. 1) A collection of dimes and quaters is worth S15.25. There are 103 coins in all. How many of each is there? 2) A collection of half dollars and nickels is worth S13.40. There are 34 coins in all. How many are there? 3) The attendance at a school concert was 578. Admission was S2.00 for adults and S1.50 for children. The total receipts were S985.00. How many adults and how many children attended? 4) A purse contains S3.90 made up of dimes and quarters. If there are 21 coins in all, how many dimes and how many quarters were there? 5) A boy has S2.25 in nickels and dimes. If there are twice as many dimes as nickels, how many of each kind has he? 6) S3.75 is made up of quarters and half dollars. If the number of quarters exceeds the number of half dollars by 3, how many coins of each denomination are there? 7) A collection of 27 coins consisting of nickels and dimes amounts to S2.25. How many coins of each kind are there? 8) S3.25 in dimes and nickels, were distributed among 45 boys. If each received one coin, how many received dimes and how many received nickels? 9) There were 429 people at a play. Admission was S1 each for adults and 75 cents each for children. The receipts were S372.50. How many children and how many adults attended? 10) There were 200 tickets sold for a women’s basketball game. Tickets for students were 50 cents each and for adults 75 cents each. The total amount of money
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 local ball game the hotdogs sold for S2.50 each and the hamburgers sold for S2.75 each. There were 131 total sandwiches sold for a total value of S342. How many of each sandwich was sold? 13) At a recent Vikings game S445 in admission tickets was taken in. The cost of a student ticket was S1.50 and the cost of a non-student ticket was S2.50. A total of 232 tickets were sold. How many students and how many nonstudents attented the game? 164 14) A bank contains 27 coins in dimes and quarters. The coins have a total value of S4.95. Find the number of dimes and quarters in the bank. 15) A coin purse contains 18 coins in nickels and dimes. The coins have a total value of S1.15. Find the number of nickels and dimes in the coin purse. 16) A business executive bought 40 stamps for S9.60. The purchase included 25c stamps and 20c stamps. How many of each type of stamp were bought? 17) A postal clerk sold some 15c stamps and some 25c stamps. Altogether, 15 stamps were sold for a total cost of S3.15. How many of each type of stamps were sold? 18) A drawer contains 15c stamps and 18c stamps. The number of 15c stamps is four less than three times the number of 18c stamps. The total value of all the stamps is S1.29. How many 15c stamps are in the drawer? 19) The total value of dimes and quarters in a bank is S6.05. There are six more quarters than dimes. Find the number of each type of coin in the bank. 20) A child’s piggy bank contains 44 coins in quarters and dimes. The coins have a total value of S8.60. Find the number of quaters in the bank. 21) A coin bank contains nickels and dimes. The number of dimes is 10 less than twice the number
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 ﬁve dollar bills. The total amount of cash in the box is S50. Find the number of each type of bill in the cash box. 23) A bank teller cashed a check for S200 using twenty dollar bills and ten dollar bills. In all, twelve bills were handed to the customer. Find the number of twenty dollar bills and the number of ten dollar bills. 24) A collection of stamps consists of 22c stamps and 40c stamps. The number of 22c stamps is three more than four times the number of 40c stamps. The total value of the stamps is S8.34. Find the number of 22c stamps in the collection. 25) A total of S27000 is invested, part of it at 12% and the rest at 13%. The total interest after one year is S3385. How much was invested at each rate? 26) A total of S50000 is invested, part of it at 5% and the rest at 7.5%. The total interest after one year is S3250. How much was invested at each rate? 27) A total of S9000 is invested, part of it at 10% and the rest at 12%. The total interest after one year is S1030. How much was invested at each rate? 28) A total of S18000 is invested, part of it at 6% and the rest at 9%. The total interest after one year is S1248. How much was invested at each rate? 29) An inheritance of S10000 is invested in 2 ways, part at 9.5% and the remainder at 11%. The combined annual interest was S1038.50. How much was invested at each rate? 165 30) Kerry earned a total of S900 last year on his investments. If S7000 was invested at a certain rate of return and S9000 was invested in a fund with a rate that was 2% higher, ﬁnd the two rates of interest. 31) Jason earned S256 interest last year on his investments. If S1600 was invested at a certain rate of return and S2400 was invested in a fund with a rate that was double the rate of the ﬁr
st fund, ﬁnd 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, ﬁnd 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 interest after one year is S385. How much was invested at each rate? 34) A total of S12000 was invested, part of it at 9% and the rest at 7.5%. The total interest after one year is S1005. How much was invested at each rate? 35) A total of S15000 is invested, part of it at 8% and the rest at 11%. The total interest after one year is S1455. How much was invested at each rate? 36) A total of S17500 is invested, part of it at 7.25% and the rest at 6.5%. The total interest after one year is S1227.50. How much was invested at each rate? 37) A total of S6000 is invested, part of it at 4.25% and the rest at 5.75%. The total interest after one year is S300. How much was invested at each rate? 38) A total of S14000 is invested, part of it at 5.5% and the rest at 9%. The total interest after one year is S910. How much was invested at each rate? 39) A total of S11000 is invested, part of it at 6.8% and the rest at 8.2%. The total interest after one year is S797. How much was invested at each rate? 40) An investment portfolio earned S2010 in interest last year. If S3000 was invested at a certain rate of return and S24000 was invested in a fund with a rate that was 4% lower, ﬁnd the two rates of interest. 41) Samantha earned S1480 in interest last year on her investments. If S5000 was invested at a certain rate of return and S11000 was invested in a fund with a rate that was two-thirds the rate of the ﬁrst fund, ﬁnd the two rates of interest. 42) A man has S5.10 in nickels, dimes, and quarters. There are
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 having a value of S3.75. If there are 40 coins in all and 3 times as many dimes as quarters, how many coins of each kind were there? 166 4.6 Systems of Equations - Mixture Problems Objective: Solve mixture problems by setting up a system of equations. One application of systems of equations are mixture problems. Mixture problems are ones where two diﬀerent solutions are mixed together resulting in a new ﬁnal solution. We will use the following table to help us solve mixture problems: Amount Part Total Item 1 Item 2 Final The ﬁrst column is for the amount of each item we have. The second column is labeled “part”. If we mix percentages we will put the rate (written as a decimal) in this column. If we mix prices we will put prices in this column. Then we can multiply the amount by the part to ﬁnd the total. Then we can get an equation by adding the amount and/or total columns that will help us solve the problem and answer the questions. These problems can have either one or two variables. We will start with one variable problems. Example 191. A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the ﬁnal solution is 60% methane? Amount Part Total Start Add Final 70 x 0.5 0.8 Set up the mixture table. We start with 70, but don′t know how much we add, that is x. The part is the percentages, 0.5 for start, 0.8 for add. 167 Amount Part Total Start Add Final 70 x 70 + x 0.5 0.8 0.6 Amount Part 0.5 0.8 0.6 70 x 70 + x Total 35 0.8x 42 + 0.6x Start Add Final Add amount column to get ﬁnal amount. The part for this amount is 0.6 because we want the ﬁnal solution to be 60% methane. Multiply amount by
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 − − 35 + 0.2x = 42 35 35 − 0.2x = 7 0.2 0.2 − Subtract 35 from both sides Divide both sides by 0.2 x = 35 We have our x! 35 mL must be added Our Solution The same process can be used if the starting and ﬁnal amount have a price attached to them, rather than a percentage. Example 192. A coﬀee mix is to be made that sells for S2.50 by mixing two types of coﬀee. The cafe has 40 mL of coﬀee that costs S3.00. How much of another coﬀee that costs S1.50 should the cafe mix with the ﬁrst? Amount Part Total Start Add Final 40 x 3 1.5 Amount Part Total Start Add Final 40 x 40 + x 3 1.5 2.5 Set up mixture table. We know the starting amount and its cost, S3. The added amount we do not know but we do know its cost is S1.50. Add the amounts to get the ﬁnal amount. We want this ﬁnal amount to sell for S2.50. 168 Amount Part Start Add Final 40 x 40 + x 3 1.5 2.5 Total 120 1.5x 100 + 2.5x Multiply amount by part to get the total. Be sure to distribute on the last row (40 + x)2.5 120 + 1.5x = 100 + 2.5x Adding down the total column gives our equation 1.5x Move variables to one side by subtracting 1.5x 1.5x − − 120 = 100 + x 100 100 20 = x We have our x. − − Subtract 100 from both sides 20mL must be added. Our Solution World View Note: Brazil is the world’s largest coﬀee producer, producing 2.59 million metric tons of coﬀee a year! That is over three times as much coﬀee as second place Vietnam! The above problems illustrate how we can put
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 to end up with 42 gallons of 20% butterfat? Amount Part Total Milk 1 Milk 2 Final x y 42 0.24 0.18 0.2 Amount Part Total 0.24 0.24x 0.18 0.18y 8.4 0.2 x y 42 Milk 1 Milk 2 Final We don′t know either start value, but we do know ﬁnal is 42. Also ﬁll in part column with percentage of each type of milk including the ﬁnal solution Multiply amount by part to get totals. x + y = 42 0.24x + 0.18y = 8.4 The amount column gives one equation The total column gives a second equation. 169 − 0.18(x + y) = (42)( 0.18y = 0.18x − − 0.18) 7.56 − − − 0.18x − − 0.18y = 7.56 0.24x + 0.18y = 8.4 0.06x = 0.84 0.06 0.06 Use addition. Multiply ﬁrst equation by 0.18 − Add the equations together Divide both sides by 0.06 x = 14 We have our x, 14 gal of 24% butterfat (14) + y = 42 14 14 Plug into original equation to ﬁnd y Subtract 14 from both sides y = 28 We have our y, 28 gal of 18% butterfat − − 14 gal of 24% and 28 gal of 18% Our Solution The same process can be used to solve mixtures of prices with two unknowns. Example 194. In a candy shop, chocolate which sells for S4 a pound is mixed with nuts which are sold for S2.50 a pound are mixed to form a chocolate-nut candy which sells for S3.50 a pound. How much of each are used to make 30 pounds of the mixture? Amount Part Total Chocolate Nut Final c n 30 4 2.5 3.5 Amount Part Total Chocolate Nut Final c n 30 4 2.5 3.5 4c 2.5n
105 Using our mixture table, use c and n for variables We do know the ﬁnal 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 ﬁrst column Second equation comes from the total column c + n = 30 We will solve this problem with substitution Solve for c by subtracting n from the ﬁrst equation n − c = 30 n n − − 170 4(30 120 − − n) + 2.5n = 105 4n + 2.5n = 105 1.5n = 105 120 120 120 15 1.5 − − Substitute into untouched equation Distribute Combine like terms Subtract 120 from both sides Divide both sides by 1.5 − 1.5n = − 1.5 − n = 10 We have our n, 10 lbs of nuts − − − c = 30 (10) Plug into c = equation to ﬁnd c − c = 20 We have our c, 20 lbs of chocolate 10 lbs of nuts and 20 lbs of chocolate Our Solution With mixture problems we often are mixing with a pure solution or using water which contains none of our chemical we are interested in. For pure solutions, the percentage is 100% (or 1 in the table). For water, the percentage is 0%. This is shown in the following example. Example 195. A solution of pure antifreeze is mixed with water to make a 65% antifreeze solution. How much of each should be used to make 70 L? Amount Part Final Antifreeze Water Final a w 70 1 0 0.65 We use a and w for our variables. Antifreeze is pure, 100% or 1 in our table, written as a decimal. Water has no antifreeze, its percentage is 0. We also ﬁll in the ﬁnal percent Amount Part Final Antifreeze Water Final a w 70 1 0 0.65 a 0 45.5 Multiply to ﬁnd ﬁnal amounts a + w = 70 a = 45.5 First equation comes from ﬁrst column Second equation comes from second column (45.5) + w = 70 We have a, plug into to other equation 45.5 Subtract 45.5 from both sides 45.5 − − 45.5L
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 antifreeze to make a solution that is 50% antifreeze? 3) Of 12 pounds of salt water 10% is salt; of another mixture 3% is salt. How many pounds of the second should be added to the ﬁrst in order to get a mixture of 5% salt? 4) How much alcohol must be added to 24 gallons of a 14% solution of alcohol in order to produce a 20% solution? 5) How many pounds of a 4% solution of borax must be added to 24 pounds of a 12% solution of borax to obtain a 10% solution of borax? 6) How many grams of pure acid must be added to 40 grams of a 20% acid solution to make a solution which is 36% acid? 7) A 100 LB bag of animal feed is 40% oats. How many pounds of oats must be added to this feed to produce a mixture which is 50% oats? 8) A 20 oz alloy of platinum that costs S220 per ounce is mixed with an alloy that costs S400 per ounce. How many ounces of the S400 alloy should be used to make an alloy that costs S300 per ounce? 9) How many pounds of tea that cost S4.20 per pound must be mixed with 12 lb of tea that cost S2.25 per pound to make a mixture that costs S3.40 per pound? 10) How many liters of a solvent that costs S80 per liter must be mixed with 6 L of a solvent that costs S25 per liter to make a solvent that costs S36 per liter? 11) How many kilograms of hard candy that cost S7.50 per kilogram must be mixed with 24 kg of jelly beans that cost S3.25 per kilogram to make a mixture that sells for S4.50 per kilogram? 12) How many kilograms of soil supplement that costs S7.00 per kilogram must be mixed with 20 kg of aluminum nitrate that costs S3.50 per kilogram to make a fertilizer that costs
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 make a mixture that costs S1.90 per liter? 15) Solution A is 50% acid and solution B is 80% acid. How much of each should be used to make 100cc. of a solution that is 68% acid? 172 16) A certain grade of milk contains 10% butter fat and a certain grade of cream 60% butter fat. How many quarts of each must be taken so as to obtain a mixture of 100 quarts that will be 45% butter fat? 17) A farmer has some cream which is 21% butterfat and some which is 15% butter fat. How many gallons of each must be mixed to produce 60 gallons of cream which is 19% butterfat? 18) A syrup manufacturer has some pure maple syrup and some which is 85% maple syrup. How many liters of each should be mixed to make 150L which is 96% maple syrup? 19) A chemist wants to make 50ml of a 16% acid solution by mixing a 13% acid solution and an 18% acid solution. How many milliliters of each solution should the chemist use? 20) A hair dye is made by blending 7% hydrogen peroxide solution and a 4% hydrogen peroxide solution. How many mililiters of each are used to make a 300 ml solution that is 5% hydrogen peroxide? 21) A paint that contains 21% green dye is mixed with a paint that contains 15% green dye. How many gallons of each must be used to make 60 gal of paint that is 19% green dye? 22) A candy mix sells for S2.20 per kilogram. It contains chocolates worth S1.80 per kilogram and other candy worth S3.00 per kilogram. How much of each are in 15 kilograms of the mixture? 23) To make a weed and feed mixture, the Green Thumb Garden Shop mixes fertilizer worth S4.00/lb. with a weed killer worth S8.00/lb. The mixture will cost S6.00/lb. How much of each should be used to prepare 500
lb. of the mixture? 24) A grocer is mixing 40 cent per lb. coﬀee with 60 cent per lb. coﬀee to make a mixture worth 54c per lb. How much of each kind of coﬀee 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 per pound. What quantity of each must he take? 26) A high-protein diet supplement that costs S6.75 per pound is mixed with a vitamin supplement that costs S3.25 per pound. How many pounds of each should be used to make 5 lb of a mixture that costs S4.65 per pound? 27) A goldsmith combined an alloy that costs S4.30 per ounce with an alloy that costs S1.80 per ounce. How many ounces of each were used to make a mixture of 200 oz costing S2.50 per ounce? 28) A grocery store oﬀers a cheese and fruit sampler that combines cheddar cheese that costs S8 per kilogram with kiwis that cost S3 per kilogram. How many kilograms of each were used to make a 5 kg mixture that costs S4.50 per kilogram? 173 29) The manager of a garden shop mixes grass seed that is 60% rye grass with 70 lb of grass seed that is 80% rye grass to make a mixture that is 74% rye grass. How much of the 60% mixture is used? 30) How many ounces of water evaporated from 50 oz of a 12% salt solution to produce a 15% salt solution? 31) A caterer made an ice cream punch by combining fruit juice that cost S2.25 per gallon with ice cream that costs S3.25 per gallon. How many gallons of each were used to make 100 gal of punch costing S2.50 per pound? 32) A clothing manufacturer has some pure silk thread and some thread that is 85% silk. How many kilograms of each must be woven together to make 75 kg of cloth that is 96% silk? 33) A carpet manufacturer blends two ﬁbers, one 20% wool and the second 50% wool. How many pounds of each ﬁber should be woven together to produce 600 lb of a fabric that is 28% wool? 34) How many pounds of coﬀee that is 40% java
beans must be mixed with 80 lb of coﬀee that is 30% java beans to make a coﬀee 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? 36) A tea that is 20% jasmine is blended with a tea that is 15% jasmine. How many pounds of each tea are used to make 5 lb of tea that is 18% jasmine? 37) How many ounces of dried apricots must be added to 18 oz of a snack mix that contains 20% dried apricots to make a mixture that is 25% dried apricots? 38) How many mililiters of pure chocolate must be added to 150 ml of chocolate topping that is 50% chocolate to make a topping that is 75% chocolate? 39) How many ounces of pure bran ﬂakes must be added to 50 oz of cereal that is 40% bran ﬂakes to produce a mixture that is 50% bran ﬂakes? 40) A ground meat mixture is formed by combining meat that costs S2.20 per pound with meat that costs S4.20 per pound. How many pounds of each were used to make a 50 lb mixture tha costs S3.00 per pound? 41) How many grams of pure water must be added to 50 g of pure acid to make a solution that is 40% acid? 42) A lumber company combined oak wood chips that cost S3.10 per pound with pine wood chips that cost S2.50 per pound. How many pounds of each were used to make an 80 lb mixture costing S2.65 per pound? 43) How many ounces of pure water must be added to 50 oz of a 15% saline solution to make a saline solution that is 10% salt? 174 175 Chapter 5 : Polynomials 5.1 Exponent Properties..............................................................................177 5.2 Negative Exponents...............................................................................183 5.3 Scientiﬁc Notation.................................................................................188 5.4 Introduction to Polynomials..................................................................192 5.5 Multiply Polynomials.............................................................................196 5.6 Multiply Special Products.....................................................................201 5.7 Divide Polynomials................................................................................205 176
5.1 Polynomials - Exponent Properties Objective: Simplify expressions using the properties of exponents. Problems with expoenents can often be simpliﬁed using a few basic exponent properties. Exponents represent repeated multiplication. We will use this fact to discover the important properties. World View Note: The word exponent comes from the Latin “expo” meaning out of and “ponere” meaning place. While there is some debate, it seems that the Babylonians living in Iraq were the ﬁrst to do work with exponents (dating back to the 23rd century BC or earlier!) Example 196. a3a2 Expand exponents to multiplication problem (aaa)(aa) Now we have 5a′s being multiplied together a5 Our Solution A quicker method to arrive at our answer would have been to just add the exponents: a3a2 = a3+2 = a5 This is known as the product rule of exponents Product Rule of Exponents: aman = am+n The product rule of exponents can be used to simplify many problems. We will add the exponent on like variables. This is shown in the following examples Example 197. Example 198. 32 36 · Same base, add the exponents 2 + 6 + 1 3 · 39 Our Solution 2x3y5z 5xy2z3 Multiply 2 · 10x4y7z4 Our Solution · 5, add exponents on x, y and z Rather than multiplying, we will now try to divide with exponents Example 199. a5 a2 aaaaa aa aaa Expand exponents Divide out two of the a′s Convert to exponents a3 Our Solution 177 A quicker method to arrive at the solution would have been to just subtract the exponents, a5 2 = a3. This is known as the quotient rule of exponents. a2 = a5 − Quotient Rule of Exponents: am an = am−n The quotient rule of exponents can similarly be used to simplify exponent problems by subtracting exponents on like variables. This is shown in the following examples. Example 200. Example 201. Same base, subtract the exponents 713 75 78 Our Solution 5a3b5c2 2ab3c Subtract exponents on a, b and c 5 2 a2b2c Our Solution A third property we will look at will have an exponent problem raised to a second exponent.
This is investigated in the following example. Example 202. 3 a2 This means we have a2 three times a2 a2 · · 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. · Power of a Power Rule of Exponents: (am)n = amn This property is often combined with two other properties which we will investigate now. Example 203. (ab)3 This means we have (ab) three times (ab)(ab)(ab) Three a′s and three b′s can be written with exponents a3b3 Our Solution 178 A quicker method to arrive at the solution would have been to take the exponent of three and put it on each factor in parenthesis, (ab)3 = a3b3. This is known as the power of a product rule or exponents. Power of a Product Rule of Exponents: (ab)m = ambm It is important to be careful to only use the power of a product rule with multiplication inside parenthesis. This property does NOT work if there is addition or subtraction. Warning 204. (a + b)m am + bm These are NOT equal, beware of this error! Another property that is very similar to the power of a product rule is considered next. Example 205 This means we have the fraction three timse Multiply fractions across the top and bottom, using exponents a3 b3 Our Solution A quicker method to arrive at the solution would have been to put the exponent on every factor in both the numerator and denominator, b3. This is known as the power of a quotient rule of exponents. = a3 a b 3 Power of a Quotient Rule of Exponents: a b m = am bm The power of a power, product and quotient rules are often used together to simplify expressions. This is shown in the following examples. Example 206. (x3yz2)4 Put the exponent of 4 on each factor, multiplying powers x12y4z8 Our solution 179 Example 207. a3b c8d5 2 Put the exponent of 2 on each factor, multiplying powers a6b2 c8d10 Our Solution As we multiply exponents its important to remember these properties apply to exponents, not bases. An expressions
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. × × 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 get 12, this would have been incorrect. Never multipy a base by the exponent. These properties pertain to exponents only, not bases. In this lesson we have discussed 5 diﬀerent exponent properties. These rules are summarized in the following table. Rules of Exponents Product Rule of Exponents Quotient Rule of Exponents aman = am+n am an = am−n (am)n = amn Power of a Power Rule of Exponents Power of a Product Rule of Exponents (ab)m = ambm Power of a Quotient Rule of Exponents a b m = am bm These ﬁve properties are often mixed up in the same problem. Often there is a bit of ﬂexibility as to which property is used ﬁrst. However, order of operations still applies to a problem. For this reason it is the suggestion of the auther to simplify inside any parenthesis ﬁrst, then simplify any exponents (using power rules), and ﬁnally simplify any multiplication or division (using product and quotient rules). This is illustrated in the next few examples. Example 209. (4x3y 5x4y2)3 In parenthesis simplify using product rule, adding exponents · (20x7y3)3 With power rules, put three on each factor, multiplying exponents 203x21y9 Evaluate 203 8000x21y9 Our Solution 180 Example 210. 7a3(2a4)3 Parenthesis are already simpliﬁed, next use power rules 7a3(8a12) Using product rule, add exponents and multiply numbers 56a15 Our Solution Example 211. 3a3b 10a4b3 · 2a4b2 Simplify numerator with product rule, adding exponents 30a7b4 2a4b2 Now use the quotient rule to subtract exponents 15a3b2 Our Solution Example 212. 3m8n12 (m2n
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 ﬁrst, using power rule in numerator Simplify numerator using product rule Simplify using the quotient rule 4a8b)2 Now that the parenthesis are simpliﬁed, use the power rules 16a16b2 Our Solution Clearly these problems can quickly become quite involved. Remember to follow order of operations as a guide, simplify inside parenthesis ﬁrst, then power rules, then product and quotient rules. 181 5.1 Practice - Exponent Properties Simplify. 44 · 2) 4 42 44 · · 44 22 1) 4 · 3) 4 · 5) 3m 4mn · 7) 2m4n2 4nm2 · 9) (33)4 11) (44)2 13) (2u3v2)2 15) (2a4)4 17) 45 43 19) 32 3 21) 3nm2 3n 23) 4x3y4 3xy3 25) (x3y4 2x2y3)2 · 27) 2x(x4y4)4 29) 31) 33) 35) 2x 7y5 3x3y 4x2y3 · (2x)3 x3 2 2y17 (2x2y4)4 3 2m4n4 2m n4 · mn4 37) 2xy5 · 2xy4 2x2y3 y3 · 3 39) q3r2 · (2p2q2r3)2 2p3 41) zy3 z3 x4y4 4 · x3y3z3 43) 2x2y2z6 · (x2z3)2 2zx2y2 4) 3 · 6) 3x 33 32 · 4x2 · 8) x2 y4 · 10) (43)4 xy2 12) (32)3 14) (xy)3 16) (2xy)4 18) 37 33 20) 34 3 22) x2y4 4xy 24) x
y3 4xy 26) (u2v2 2u4)3 · 2v3 · 2u3v 2b4 · 3a3b4 · 28) 3vu5 uv2 30) 2ba7 ba2 32) 2a2b2a7 (ba4)2 · 34) yx2 (y4)2 · 2y4 36) n3(n4)2 2mn 38) (2y 3x2)2 x2 2x2y4 · 40) 2x4y5 · 2z10 x2y7 (xy2z2)4 2q3 p3r4 2p3 · (qrp3)2 4 42) 182 5.2 Polynomials - Negative Exponents Objective: Simplify expressions with negative exponents using the properties of exponents. There are a few special exponent properties that deal with exponents that are not positive. The ﬁrst is considered in the following example, which is worded out 2 diﬀerent ways: Example 214. a3 a3 Use the quotient rule to subtract exponents a0 Our Solution, but now we consider the problem a the second way: a3 a3 Rewrite exponents as repeated multiplication aaa aaa Reduce out all the a′s 1 1 = 1 Our Solution, when we combine the two solutions we get: a0 = 1 Our ﬁnal result. This ﬁnal result is an imporant property known as the zero power rule of exponents Zero Power Rule of Exponents: a0 = 1 Any number or expression raised to the zero power will always be 1. This is illustrated in the following example. Example 215. (3x2)0 Zero power rule 1 Our Solution Another property we will consider here deals with negative exponents. Again we will solve the following example two ways. 183 Example 216. a3 a5 Using the quotient rule, subtract exponents a− 2 Our Solution, but we will also solve this problem another way. a3 a5 Rewrite exponents as repeated multiplication aaa aaaaa 1 aa Reduce three a′s out of top and bottom Simplify to exponents 1 a2 Our Solution, putting these solutions together gives: a− 2 = 1 a2 Our Final Solution This example illustrates an important property of exponents. Negative exponents yield the reciprocal of the base. Once we take the reciprical the
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: a−m = 1 m 1 a−m = am a b −m = bm am Negative exponents can be combined in several diﬀerent ways. As a general rule if we think of our expression as a fraction, negative exponents in the numerator must be moved to the denominator, likewise, negative exponents in the denominator need to be moved to the numerator. When the base with exponent moves, the exponent is now positive. This is illustrated in the following example. Example 217. 2c a3b− 4f 2 Negative exponents on b, d, and e need to ﬂip 1 e− 2d− a 3cde4 2b2f 2 Our Solution 184 As we simpliﬁed our fraction we took special care to move the bases that had a negative exponent, but the expression itself did not become negative because of those exponents. Also, it is important to remember that exponents only eﬀect what they are attached to. The 2 in the denominator of the above example does not have an exponent on it, so it does not move with the d. We now have the following nine properties of exponents. It is important that we are very familiar with all of them. Properties of Exponents aman = am+n (ab)m = ambm am an = am−n (am)n = amn a b m = am bm a0 = 1 a−m = 1 am 1 a−m = am a b −m = bm am World View Note: Nicolas Chuquet, the French mathematician of the 15th century wrote 121m¯ to indicate 12x− 1. This was the ﬁrst known use of the negative exponent. Simplifying with negative exponents is much the same as simplifying with positive exponents. It is the advice of the author to keep the negative exponents until the end of the problem and then move them around to their correct location (numerator or denominator). As we do this it is important to be very careful of rules for adding, subtracting, and multiplying with negatives. This is illustrated in the following examples Example 218. 4x− 2 3 5y− 6x−
3x3y− · 5y3 5 12x− 6x− 2y− 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 ( − 3 = ( − 5 2x3y− 2x3 y8 Our Solution 185 Example 219. 3 (3ab3)− 2a− 2ab− 4b0 In numerator, use power rule with In denominator, b0 = 1 − 2, multiplying exponents 3− 3 2a− 2b− 6ab− 2a− 4 3− 9 2a− 2a− 1b− 4 3− 9 2a3b− 2 a3 322b9 In numerator, use product rule to add exponents Use quotient rule to subtract exponents, be careful with negatives 1) ( − − ( − 4) = ( − 1) + 4 = 3 Move 3 and b to denominator because of negative exponents Evaluate 322 a3 18b9 Our Solution In the previous example it is important to point out that when we simpliﬁed 3− we moved the three to the denominator and the exponent became positive. We did not make the number negative! Negative exponents never make the bases negative, they simply mean we have to take the reciprocal of the base. One ﬁnal example with negative exponents is given here. 2 Example 220. 3x− 2y5z3 6x− · 9(x2y− 2)− 6y− 3 2z − 3 3 − 3 − 18x− 9x− 8y3z0 6y6 In numerator, use product rule, adding exponents In denominator, use power rule, multiplying exponets Use quotient rule to subtract exponents, be careful with negatives: ( 3 ( 8) − − 6 = 3 + ( 6) = ( 8) + 6 = 2 − − − 6) = 3 Parenthesis are done, use power rule with 3z0)− 3x6y9z0 Move 2 with negative exponent down and z0 = 1 − − − − 3 3 2y− 2− (2x− x6y9 23 x6y9 8 Evaluate 23 Our Solution 186 5.2 Practice - Negative
Exponents Simplify. Your answer should contain only positive expontents. 1) 2x4y− 2 3) (a4b− · 3)3 (2xy3)4 2a3b− 2 · 5) (2x2y2)4x− 4 · 2x−3y2 3x0 9) 3x−3y3 11) 4xy−3 · x−4y0 · 4y−1 2) 2a− 3 2b− (2a0b4)4 · (2x3)0 4) 2x3y2 · 6) (m0n3 2m− 3n− 3)0 · 1n− 3y3 2x4y−3 · 3x3y2 10) 3yx3 12) 4y −2 · 3x−2y−4 (2m− 1n− 3)4 3 · 7) (x3y4)3 x− 4y4 8) 2m− 13) 15) u2v−1 2u0v4 · u2 4u0v3 · 2uv 3v2 17) 2y (x0y2)4 19) ( 2a2b3 a−1 )4 2nm4 (2m2n2)4 21) 23) (2mn)4 m0n−2 25) y3 x−3y2 (x4y2)3 · 27) 2u−2v3 · 2u−4v0 (2uv4)−1 29) ( 2x0 · y4 y4 )3 31) y(2x4y2)2 2x4y0 33) 2x4y4z−2 (zy2)4 2yzx2 · 2h−3k0 35) 2kh0 · (2kj3)2 37) (cb3)2 · (a3b−2c3)3 2a−3b2 4x3y−4 4x · 14) 2xy2 · 4x−4y−4 16) 2x−2y2 4yx2 18) (a4)4 2b 20) ( 2y −4 x2 )− 2 22) 2y2 (x4y0)−4 24) 2x−3 (x4y−3)−1 26) 2x−2y0 · (
xy0)−1 2xy4 28) 2yx2 x−2 · (2x0y4)−1 30) u−3v−4 2v(2u−3v4)0 b−1 32) (2a4b0)0 34) 2b4c−2 · a−2b4 2a−3b2 · (2b3c2)−4 36) ( (2x−3y0z−1)3 2x3 · x−3y2 2 )− 38) 2q4 m2p2q4 (2m−4p2)3 · 39) (yx−4z2)−1 x2y3z−1 z3 · 40) 2mpn−3 (m0n−4p2)3 2n2p0 · 187 5.3 Polynomials - Scientiﬁc Notation Objective: Multiply and divide expressions using scientiﬁc notation and exponent properties. One application of exponent properties comes from scientiﬁc notation. Scientiﬁc notation is used to represent really large or really small numbers. An example of really large numbers would be the distance that light travels in a year in miles. An example of really small numbers would be the mass of a single hydrogen atom in grams. Doing basic operations such as multiplication and division with these numbers would normally be very combersome. However, our exponent properties make this process much simpler. First we will take a look at what scientiﬁc notation is. Scientiﬁc notation has two parts, a number between one and ten (it can be equal to one, but not ten), and that number multiplied by ten to some exponent. Scientiﬁc Notation: a 10b where 1 6 a < 10 × The exponent, b, is very important to how we convert between scientiﬁc notation and normal numbers, or standard notation. The exponent tells us how many times we will multiply by 10. Multiplying by 10 in aﬀect moves the decimal point one place. So the exponent will tell us how many times the exponent moves between scientiﬁc notation and standard notation. To decide which direction to move the decimal (left or right) we simply need to remember that positive exponents mean in standard notation we have a big number (bigger than ten)
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 scientiﬁc notation. Example 221. Convert 14, 200 to scientiﬁc notation Put decimal after ﬁrst nonzero number 1.42 Exponent is how many times decimal moved, 4 104 Positive exponent, standard notation is big 104 Our Solution × × 1.42 Example 222. Convert 0.0042 to scientiﬁc notation Put decimal after ﬁrst nonzero number 4.2 Exponent is how many times decimal moved, 3 3 Negative exponent, standard notation is small 3 Our Solution 10− 10− 4.2 × × 188 Example 223. Convert 3.21 × 105 to standard notation Positive exponent means standard notation big number. Move decimal right 5 places 321, 000 Our Solution Example 224. Conver 7.4 10− × 3 to standard notation Negative exponent means standard notation is a small number. Move decimal left 3 places 0.0074 Our Solution Converting between standard notation and scientiﬁc notation is important to understand how scientiﬁc notation works and what it does. Here our main interest is to be able to multiply and divide numbers in scientiﬁc notation using exponent properties. The way we do this is ﬁrst do the operation with the front number (multiply or divide) then use exponent properties to simplify the 10’s. Scientiﬁc notation is the only time where it will be allowed to have negative exponents in our ﬁnal solution. The negative exponent simply informs us that we are dealing with small numbers. Consider the following examples. Example 225. (2.1 × 7)(3.7 10− (2.1)(3.7) = 7.77 Multiply numbers × 105) Deal with numbers and 10′s separately 10− 7105 = 10− 10− 7.77 2 Use product rule on 10′s and add exponents 2 Our Solution × Example 226. 3 Deal with numbers and 10′s separately = 1.6 Divide Numbers 4.96 3.1 104 × 10− × 4.96 3.1 104 10− 3 = 107 Use quotient rule to subtract exponents, be careful with negatives! Be careful with negatives, 4 ( − − 3) = 4 + 3 = 7 107 Our Solution 1.
6 × 189 Example 227. 4)3 Use power rule to deal with numbers and 10′s separately × 10− (1.8 1.83 = 5.832 Evaluate 1.83 4)3 = 10− 10− 12 Multiply exponents 12 Our Solution 5.832 (10− × Often when we multiply or divide in scientiﬁc notation the end result is not in scientiﬁc notation. We will then have to convert the front number into scientiﬁc notation and then combine the 10’s using the product property of exponents and adding the exponents. This is shown in the following examples. Example 228. (4.7 10− 3)(6.1 109) Deal with numbers and 10′s separately × (4.7)(6.1) = 28.67 Multiply numbers × 10110− × 101 Convert this number into scientiﬁc notation 2.867 3109 = 107 Use product rule, add exponents, using 101 from conversion 2.867 107 Our Solution × World View Note: Archimedes (287 BC - 212 BC), the Greek mathematician, developed a system for representing large numbers using a system very similar to scientiﬁc notation. He used his system to calculate the number of grains of sand it would take to ﬁll the universe. His conclusion was 1063 grains of sand because he ﬁgured the universe to have a diameter of 1014 stadia or about 2 light years. Example 229. 3 Deal with numbers and 10′s separately = 0.53 Divide numbers 2.014 3.8 10− × 7 10− × 2.014 3.8 0.53 = 5.3 10−110− 10− × 3 10− 1 Change this number into scientiﬁc notation 7 = 103 Use product and quotient rule, using 10− 1 from the conversion 5.3 × Be careful with signs: ( 1) + ( ( 3) − 103 Our Solution − − − 7) = ( 1) + ( − − 3) + 7 = 3 190 5.3 Practice - Scientiﬁc Notation Write each number in scientiﬁc notiation 1) 885 3) 0.081 5) 0.039 Write each number in standard notation 7) 8.7 x 105 9) 9 x 10− 4 11) 2 x 100
2) 0.000744 4) 1.09 6) 15000 8) 2.56 x 102 10) 5 x 104 12) 6 x 10− 5 Simplify. Write each answer in scientiﬁc notation. 13) (7 x 10− 1)(2 x 10− 3) 15) (5.26 x 10− 5)(3.16 x 10− 2) 17) (2.6 x 10− 2)(6 x 10− 2) 101 19) 4.9 × 10−3 2.7 × 21) 5.33 9.62 × × 23) (5.5 10−6 10−2 25) (7.8 10− 5)2 10− 2)5 × × 27) (8.03 104)− 4 × 10−6 10−4 29) 6.1 5.1 × × 31) (3.6 33) (1.8 × 104 10−2 10−3 × 10−6 35) 9 7.83 × × 37) 3.22 7 × 39) 2.4 6.5 41) 6 5.8 10−6 100 × × 103 10−3 × × 100)(6.1 10− 3) × × 10− 3 5)− 14) (2 10− 6)(8.8 5) 10− × × 106)(9.84 10− 1) × 16) (5.1 × 104 18) 7.4 × 10−4 1.7 × 20) 7.2 7.32 10−1 10−1 × × 10−3 100 22) 3.2 × 5.02 × 24) (9.6 26) (5.4 × 28) (6.88 × 105 × 10−2 30) 8.4 7 × 32) (3.15 × 103)− 4 106)− 3 34) 9.58 1.14 × × 36) (8.3 38) 5 × 6.69 101)5 × 106 102 × 10− × 3 2)− 104)(6 101) × × 40) (9 42) (2 191 10− 4)(4.23 101) × 10− 1) × 103)(8 × 10−2 10−3 5.4 Polynomials - Introduction to Polynomials Objective: Evaluate, add, and subtract polynomials. Many applications in mathematics have to do with what are called polynomials. Polynom
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, such as a2 b2. A Trinomial has three terms, such as ax2 + bx + c. The term polynomial means many terms. Monomials, binomials, trinomials, and expressions with more terms all fall under the umbrella of “polynomials”. − − If we know what the variable in a polynomial represents we can replace the variable with the number and evaluate the polynomial as shown in the following example. Example 230. 2x2 4x + 6 when x = − 2( − 4)2 − 2(16) − − 4( 4( 32 + 16 + 6 Add − − 4 Replace variable x with 4 − 4) + 6 Exponents ﬁrst 4) + 6 Multiplication (we can do all terms at once) 54 Our Solution − 9 because the exponent is only attached to the 3. Also, ( It is important to be careful with negative variables and exponents. Remember the exponent only eﬀects the number it is physically attached to. This means − 32 = 3)2 = 9 because the exponent is attached to the parenthesis and eﬀects everything inside. When we replace a variable with parenthesis like in the previous example, the substi4)2 = 16 in the example. However, contuted value is in parenthesis. So the ( sider the next example. − − Example 231. − x2 + 2x + 6 when x = 3 Replace variable x with 3 − (3)2 + 2(3) + 6 Exponent only on the 3, not negative − 9 + 2(3) + 6 Multiply 9 + 6 + 6 Add − 3 Our Solution 192 World View Note: Ada Lovelace in 1842 described a Diﬀerence Engine that would be used to caluclate values of polynomials. Her work became the foundation for what would become the modern computer (the programming language Ada was named in her honor), more than 100 years after her
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 − 2x + 8) + (3x3 9x2 7x3 − − 9x2 − 2x − 11) Combine like terms 4x3 + 3x3 and 8 3 Our Solution − 11 − Generally ﬁnal answers for polynomials are written so the exponent on the variable counts down. Example 3 demonstrates this with the exponent counting down 3, 2, 1, 0 (recall x0 = 1). Subtracting polynomials is almost as fast. One extra step comes from the minus in front of the parenthesis. When we have a negative in front of parenthesis we distribute it through, changing the signs of everything inside. The same is done for the subtraction sign. Example 233. (5x2 2x + 7) − 5x2 − 2x + 7 − (3x2 + 6x 3x2 − 2x2 − 6x + 4 Combine like terms 5x2 − 8x + 11 Our Solution − 4) Distribute negative through second part 3x3, 2x − − − 6x, and 7 + 4 Addition and subtraction can also be combined into the same problem as shown in this ﬁnal example. Example 234. (2x2 4x + 3) + (5x2 6x + 1) (x2 − 2x2 − 4x + 3 + 5x2 − 6x + 1 − − 9x + 8) Distribute negative through − − x2 + 9x 6x2 x − 8 Combine like terms 4 Our Solution − − 193 5.4 Practice - Introduction to Polynomials Simplify each expression. 1) a3 a2 + 6a 21 when a = − − 2) n2 + 3n − 11 when n = 6 4 − − 7n2 + 15n − 20 when n = 2 3) n3 − 4) n3 9n2 + 23n 21 when n = 5 − − 9n2 − 5n4 5) − 6) x4 11n3 n 5 when n = − 5x3 − − x + 13 when x =
5 − 1 − − − 7) x2 + 9x + 23 when x = 3 − 8) − 9) x4 6x3 + 41x2 6x3 + x2 − − − 32x + 11 when x = 6 24 when x = 6 10) m4 + 8m3 + 14m2 + 13m + 5 when m = 6 − 11) (5p 5p4) (8p 8p4) − − 12) (7m2 + 5m3) − (6m3 5m2) 13) (3n2 + n3) − − (2n3 − 7n2) − 14) (x2 + 5x3) + (7x2 + 3x3) 15) (8n + n4) (3n − 16) (3v4 + 1) + (5 − 4n4) − v4) 17) (1 + 5p3) − 18) (6x3 + 5x) (1 8p3) − (8x + 6x3) − 19) (5n4 + 6n3) + (8 20) (8x2 + 1) (6 − − − x2 3n3 − x4) − 5n4) 194 21) (3 + b4) + (7 + 2b + b4) 22) (1 + 6r2) + (6r2 − 2 23) (8x3 + 1) (5x4 − − 3r4) − 6x3 + 2) 24) (4n4 + 6) − 25) (2a + 2a4) (4n 1 − (3a2 n4) − 5a4 + 4a) − − 26) (6v + 8v3) + (3 + 4v3 3v) 6p + 3) − − 27) (4p2 3 2p) (3p2 − − 28) (7 + 4m + 8m4) − (5m4 + 1 + 6m) − 29) (4b3 + 7b2 3) + (8 + 5b2 + b3) − 8n4) (3n + 7n4 + 7) 30) (7n + 1 − 31) (3 + 2n2 + 4n4) + (n3 − − 32) (7x2
+ 2x4 + 7x3) + (6x3 33) (n − 5n4 + 7) + (n2 − 34) (8x2 + 2x4 + 7x3) + (7x4 7n2 4n4) − 8x4 n) − 7n4 7x2) − − 7x3 + 2x2) − − 8 + 2x + 6x3) 35) (8r4 − 36) (4x3 + x − 37) (2n2 + 7n4 5r3 + 5r2) + (2r2 + 2r3 7r4 + 1) 7x2) + (x2 − 2) + (2 + 2n3 + 4n2 + 2n4) 38) (7b3 39) (8 − − 4b + 4b4) − b + 7b3) (8b3 − − (3b4 + 7b − − 8n3) + (7n4 + 2 40) (1 3n4 − − 4b2 + 2b4 8b) − 8 + 7b2) + (3 3b + 6b3) − 6n2 + 3n3) + (4n3 + 8n4 + 7) − 41) (8x4 + 2x3 + 2x) + (2x + 2 2x3 42) (6x 5x4 4x2) (2x − − − − − 7x2 − 4x4 − x4) − (x3 + 5x4 + 8x) 8) − − (8 − 6x2 − 4x4) 195 5.5 Polynomials - Multiplying Polynomials Objective: Multiply polynomials. Multiplying polynomials can take several diﬀerent forms based on what we are multiplying. We will ﬁrst look at multiplying monomials, then monomials by polynomials and ﬁnish with polynomials by polynomials. Multiplying monomials is done by multiplying the numbers or coeﬃcients and then adding the exponents on like factors. This is shown in the next example. Example 235. (4x3y4z)(2x2y6z3) Multiply numbers and add exponents for
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 when we multiply (or divide) the exponents will be changing. Next we consider multiplying a monomial by a polynomial. We have seen this operation before with distributing through parenthesis. Here we will see the exact same process. Example 236. 2x + 5) Distribute the 4x3, multiplying numbers, adding exponents 4x3(5x2 20x5 − − 8x4 + 20x3 Our Solution Following is another example with more variables. When distributing the exponents on a are added and the exponents on b are added. Example 237. 2a3b(3ab2 6a4b3 − − 4a) Distribute, multiplying numbers and adding exponents 8a4b Our Solution There are several diﬀerent methods for multiplying polynomials. All of which work, often students prefer the method they are ﬁrst taught. Here three methods will be discussed. All three methods will be used to solve the same two multiplication problems. Multiply by Distributing 196 Just as we distribute a monomial through parenthesis we can distribute an entire polynomial. As we do this we take each term of the second polynomial and put it in front of the ﬁrst polynomial. Example 238. (4x + 7y)(3x 2y) Distribute (4x + 7y) through parenthesis 3x(4x + 7y) 2y(4x + 7y) Distribute the 3x and 2y − 12x2 + 21xy 8xy − 12x2 + 13xy − 14y2 Combine like terms 21xy 14y2 Our Solution 8xy − − − − This example illustrates an important point, the negative/subtraction sign stays with the 2y. Which means on the second step the negative is also distributed through the last set of parenthesis. Multiplying by distributing can easily be extended to problems with more terms. First distribute the front parenthesis onto each term, then distribute again! Example 239. 4x2(
2x − 8x3 (2x 7x(2x − − − 5)(4x2 5) + 3(2x 14x2 + 35x + 6x 34x2 + 41x 8x3 − 5) − − 20x2 − − − − 7x + 3) Distribute (2x 5) through parenthesis − 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 procedure known as factoring. Factoring will be addressed in a future lesson. Multiply by FOIL Another form of multiplying is known as FOIL. Using the FOIL method we multiply each term in the ﬁrst binomial by each term in the second binomial. The letters of FOIL help us remember every combination. F stands for First, we multiply the ﬁrst term of each binomial. O stand for Outside, we multiply the outside two terms. I stands for Inside, we multiply the inside two terms. L stands for Last, we multiply the last term of each binomial. This is shown in the next example: Example 240. (4x + 7y)(3x (4x)(3x) = 12x2 − (4x)( 2y) = (7y)(3x) = 21xy − − 2y) Use FOIL to multiply − − − − First terms (4x)(3x) Outside terms (4x)( Inside terms (7y)(3x) Last terms (7y)( F 8xy O I 14y2 L 14y2 Combine like terms 14y2 Our Solution 2y) − 2y) − 8xy + 21xy − 12x2 (7y)( 2y) = − 8xy + 21xy 12x2 + 13xy − − − − 197 Some students like to think of the FOIL method as distributing the ﬁrst term 4x through the (3x 2y). Thinking about FOIL in this way makes it possible to extend this method to problems with more terms. 2y) and distributing the second term 7y through the (3x − − Example 241. (2x)(4x2) + (2x)( − (2x 5)(4x2 7x + 3) Distribute 2x and 5 − 7x) + (2x)(3) 8x3 − 14x2 +
6x 8x3 − − 5(4x2) 5( − 7x) − − − 20x2 + 35x 34x2 + 41x − − − − 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) − and do the multiplication mentally to come up with 12x2 14y2 and − then combine like terms to come up with the ﬁnal solution. 8xy + 21xy − Multiplying in rows A third method for multiplying polynomials looks very similar to multiplying numbers. Consider the problem: 35 27 × 245 Multiply 7 by 5 then 3 700 Use 0 for placeholder, multiply 2 by 5 then 3 945 Add to get Our Solution World View Note: The ﬁrst known system that used place values comes from Chinese mathematics, dating back to 190 AD or earlier. The same process can be done with polynomials. Multiply each term on the bottom with each term on the top. Example 242. (4x + 7y)(3x 2y) Rewrite as vertical problem − 4x + 7y 2y 3x × 8xy − 12x2 + 21xy 12x2 + 13xy − − − 14y2 Multiply 2y by 7y then 4x − Multiply 3x by 7y then 4x. Line up like terms 14y2 Add like terms to get Our Solution This same process is easily expanded to a problem with more terms. 198 Example 243. (2x − 8x3 8x3 − − − − − × 5)(4x2 4x3 7x + 3) Rewrite as vertical problem 7x + 3 Put polynomial with most terms on top 2x 20x2 + 35x 14x2 + 6x 34x2 + 41x 5 − 15 Multiply − 15 Add like terms to get our solution 5 by each term Multiply 2x by each term. Line up like terms − − This method of multiplying in rows also works with multiplying a monomial by a polynomial! Any of the three described methods work to multiply polynomials. It is suggested that you are very comfortable with at least one of these methods as you work through the practice problems
. All three methods are shown side by side in the example. Example 244. (2x − y)(4x 5y) − 4x(2x 8x2 Distribute 5y(2x 10xy y) − 4xy − − 14xy − 5y2 y) − 5y2 − − 8x2 − 2x(4x) + 2x( FOIL 5y) − 10xy − − y( y(4x) 4xy + 5y2 − 14xy + 5y2 8x2 − 8x2 − 5y) − Rows y 2x 5y 4x − × − 10xy + 5y2 4xy 14xy + 5y2 − 8x2 − 8x2 − When we are multiplying a monomial by a polynomial by a polynomial we can solve by ﬁrst multiplying the polynomials then distributing the coeﬃcient last. This is shown in the last example. Example 245. 3(2x 3(2x2 + 10x − − 4)(x + 5) Multiply the binomials, we will use FOIL 4x 3(2x2 + 6x 6x2 + 18x 20) Combine like terms 20) Distribute the 3 60 Our Solution − − − A common error students do is distribute the three at the start into both parenthesis. While we can distribute the 3 into the (2x 4) factor, distributing into both would be wrong. Be careful of this error. This is why it is suggested to multiply the binomials ﬁrst, then distribute the coeﬃenct last. − 199 5.5 Practice - Multiply Polynomials Find each product. 1) 6(p 7) − 3) 2(6x + 3) 5) 5m4(4m + 4) 7) (4n + 6)(8n + 8) 9) (8b + 3)(7b 5) − 11) (4x + 5)(2x + 3) 13) (3v 4)(5v 2) − − 15) (6x 7)(4x + 1) − 17) (5x + y)(6x 4y) − 2) 4k(8k + 4) 4) 3n2(6n + 7) 6) 3(4r 7) − 8) (2x
+ 1)(x 4) − 10) (r + 8)(4r + 8) 12) (7n − 6)(n + 7) 14) (6a + 4)(a 8) − 6)(4x 16) (5x − − 18) (2u + 3v)(8u 1) 7v) − 19) (x + 3y)(3x + 4y) 20) (8u + 6v)(5u 8v) 21) (7x + 5y)(8x + 3y) 22) (5a + 8b)(a − 3b) − 7)(6r2 r + 5) 24) (4x + 8)(4x2 + 3x + 5) 23) (r − 25) (6n − 4)(2n2 − − 27) (6x + 3y)(6x2 2n + 5) 7xy + 4y2) − 29) (8n2 + 4n + 6)(6n2 5n + 6) − 31) (5k2 + 3k + 3)(3k2 + 3k + 6) 33) 3(3x − 4)(2x + 1) 35) 3(2x + 1)(4x 5) − 37) 7(x − 39) 6(4x 5)(x 2) − 1)(4x + 1) − 26) (2b − 3)(4b2 + 4b + 4) 28) (3m − 2n)(7m2 + 6mn + 4n2) 30) (2a2 + 6a + 3)(7a2 6a + 1) − 32) (7u2 + 8uv − 6v2)(6u2 + 4uv + 3v2) 34) 5(x 4)(2x − − 36) 2(4x + 1)(2x 3) 6) − 38) 5(2x 1)(4x + 1) − 40) 3(2x + 3)(6x + 9) 200 5.6 Polynomials - Multiply Special Products Objective: Recognize and use special product rules of a sum and diﬀerence and perfect squares to multiply polynomials. There are a few shortcuts that we can take when multiplying polynomials. If we can recognize them the shortcuts can help us arrive at the solution much quicker. These shortcuts will also be
useful to us as our study of algebra continues. The ﬁrst shortcut is often called a sum and a diﬀerence. A sum and a diﬀerence is easily recognized as the numbers and variables are exactly the same, but the sign in the middle is diﬀerent (one sum, one diﬀerence). To illustrate the shortcut consider the following example, multiplied by the distributing method. Example 246. (a + b)(a b) Distribute (a + b) a(a + b) a2 + ab − − b(a + b) Distribute a and ab a2 b2 Combine like terms ab b2 Our Solution − b ab − − − − The important part of this example is the middle terms subtracted to zero. Rather than going through all this work, when we have a sum and a diﬀerence we will jump right to our solution by squaring the ﬁrst term and squaring the last term, putting a subtraction between them. This is illustrated in the following example Example 247. (x − 5)(x + 5) Recognize sum and diﬀerence x2 − 25 Square both, put subtraction between. Our Solution This is much quicker than going through the work of multiplying and combining like terms. Often students ask if they can just multiply out using another method and not learn the shortcut. These shortcuts are going to be very useful when we get to factoring polynomials, or reversing the multiplication process. For this reason it is very important to be able to recognize these shortcuts. More examples are shown here. 201 Example 248. (3x + 7)(3x 9x2 − − 7) Recognize sum and diﬀerence 49 Square both, put subtraction between. Our Solution Example 249. (2x − 6y)(2x + 6y) Recognize sum and diﬀerence 4x2 − 36y2 Square both, put subtraction between. Our Solution b2 It is interesting to note that while we can multiply and get an answer like a2 (with subtraction), it is impossible to multiply real numbers and end up with a product such as a2 + b2 (with addition). − Another shortcut used to multiply is known as a perfect square. These are easy to recognize as we will have a binomial with a 2 in the exponent. The following example illustrates multiplying a perfect square Example 250. (
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 ﬁnal parenthesis a2 + ab + ab + b2 Combine like terms ab + ab a2 + 2ab + b2 Our Solution This problem also helps us ﬁnd our shortcut for multiplying. The ﬁrst term in the answer is the square of the ﬁrst term in the problem. The middle term is 2 times the ﬁrst term times the second term. The last term is the square of the last term. This can be shortened to square the ﬁrst, twice the product, square the last. If we can remember this shortcut we can square any binomial. This is illustrated in the following example Example 251. (x − 5)2 Recognize perfect square x2 Square the ﬁrst 2(x)( 10x Twice the product − 5)2 = 25 10x + 25 Our Solution Square the last 5) = − ( x2 − − 202 Be very careful when we are squaring a binomial to NOT distribute the square 5)2 = x2 through the parenthesis. A common error is to do the following: (x 25 (or x2 + 25). Notice both of these are missing the middle term, 10x. This is why it is important to use the shortcut to help us ﬁnd the correct solution. Another important observation is that the middle term in the solution always has the same sign as the middle term in the problem. This is illustrated in the next examples. − − − Example 252. Example 253. (2x + 5)2 Recognize perfect square Square the ﬁrst (2x)2 = 4x2 2(2x)(5) = 20x Twice the product 52 = 25 Square the last 4x2 + 20x + 25 Our Solution (3x 7y)2 Recognize perfect square − 42xy + 49y2 9x2 − Square the ﬁrst, twice the product, square the last. Our Solution Example 254. (5a + 9b)2 Recognize perfect square 25a2 + 90ab + 81b2 Square the ﬁrst, twice the product, square the last. Our Solution These two formulas will be important to commit
to memory. The more familiar we are with them, the easier factoring, or multiplying in reverse, will be. The ﬁnal example covers both types of problems (two perfect squares, one positive, one negative), be sure to notice the diﬀerence between the examples and how each formula is used Example 255. (4x 7)(4x + 7) 49 − 16x2 − (4x + 7)2 16x2 + 56x + 49 (4x 7)2 − 56x + 49 16x2 − World View Note: There are also formulas for higher powers of binomials as well, such as (a + b)3 = a3 + 3a2b + 3ab2 + b3. While French mathematician Blaise Pascal often gets credit for working with these expansions of binomials in the 17th century, Chinese mathematicians had been working with them almost 400 years earlier! 203 5.6 Practice - Multiply Special Products Find each product. 1) (x + 8)(x 8) − 3) (1 + 3p)(1 3p) − 7n)(1 + 7n) 5) (1 − 7) (5n 8)(5n + 8) − 9) (4x + 8)(4x 8) − x)(4y + x) 11) (4y − 13) (4m − 8n)(4m + 8n) 15) (6x − 2y)(6x + 2y) 17) (a + 5)2 19) (x 8)2 − 21) (p + 7)2 23) (7 − 25) (5m 5n)2 8)2 − 27) (5x + 7y)2 29) (2x + 2y)2 31) (5 + 2r)2 33) (2 + 5x)2 35) (4v − 7) (4v + 7) 37) (n 5)(n + 5) − 39) (4k + 2)2 2) (a 4) (x − − 4)(a + 4) 3)(x + 3) 6) (8m + 5)(8m 5) − 3) 8) (2r + 3)(2r − 7)(b + 7) 10) (b − 12) (7a + 7b)(7a 7b) − 3x)(3y +
3x) 14) (3y − 16) (1 + 5n)2 18) (v + 4)2 20) (1 − 22) (7k 6n)2 − 7)2 5)2 24) (4x − 26) (3a + 3b)2 28) (4m n)2 − 30) (8x + 5y)2 32) (m 7)2 − 34) (8n + 7)(8n 36) (b + 4)(b − 38) (7x + 7)2 7) − 4) 40) (3a − 8)(3a + 8) 204 5.7 Polynomials - Divide Polynomials Objective: Divide polynomials using long division. Dividing polynomials is a process very similar to long division of whole numbers. But before we look at that, we will ﬁrst want to be able to master dividing a polynomial by a monomial. The way we do this is very similar to distributing, but the operation we distribute is the division, dividing each term by the monomial and reducing the resulting expression. This is shown in the following examples Example 256. 9x5 + 6x4 18x3 − 3x2 − 24x2 9x5 3x2 + 6x4 3x2 − 18x3 3x2 − 24x2 3x2 Divide each term in the numerator by 3x2 Reduce each fraction, subtracting exponents 3x3 + 2x2 6x − − 8 Our Solution Example 257. 8x3 + 4x2 − 4x2 2x + 6 Divide each term in the numerator by 4x2 8x3 4x2 + 4x2 4x2 − 2x 4x2 + 6 4x2 Reduce each fraction, subtracting exponents Remember negative exponents are moved to denominator 2x + 1 1 2x + 3 2x2 Our Solution − The previous example illustrates that sometimes we will have fractions in our solution, as long as they are reduced this will be correct for our solution. Also interesting in this problem is the second term 4x2 4x2 divided out completely. Remember that this means the reduced answer is 1 not 0. Long division is required when we divide by more than just a monomial. Long division with polynomials works very similar to long division with whole numbers. 205 An example is given to
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 − 23 Change the sign of this number (make it subtract) and combine Bring down next number 4 = 4 · 15 Repeat, divide front numbers: 23 4 = 5 4 − − 631 \| 4 23 20 31 Bring down next number Multiply this number by divisor: 5 Change the sign of this number (make it subtract) and combine 4 = 20 · 157 Repeat, divide front numbers: = 7 31 4 4 \| − − 631 4 23 20 31 Multiply this number by divisor: 7 28 Change the sign of this number (make it subtract) and combine 4 = 28 · − 3 We will write our remainder as a fraction, over the divisor, added to the end 157 3 4 Our Solution This same process will be used to multiply polynomials. The only diﬀerence is we will replace the word “number” with the word “term” Dividing Polynomials 1. Divide front terms 2. Multiply this term by the divisor 206 3. Change the sign of the terms and combine 4. Bring down the next term 5. Repeat Step number 3 tends to be the one that students skip, not changing the signs of the terms would be equivalent to adding instead of subtracting on long division with whole numbers. Be sure not to miss this step! This process is illustrated in the following two examples. Example 259. 3x3 − 5x2 x 32x + 7 − 4 − Rewrite problem as long division x 4 \| − 3x3 − 5x2 − 32x + 7 Divide front terms: 3x3 x = 3x2 x 4 − − 5x2 3x2 3x3 − − \| 3x3 + 12x2 7x2 − 32x + 7 Multiply this term by divisor: 3x2(x 32x Change the signs and combine Bring down the next term 4) = 3x3 12x2 − − 3x2 + 7x Repeat, divide front terms: 7x2 x = 7x Multiply this term by divisor: 7x(x Change the signs and combine − 4) = 7x2 28x −
4x + 7 Bring down the next term − 4x Repeat, divide front terms: − x = 4 − x − 4 \| − 32x + 7 5x2 3x3 − − 3x3 + 12x2 7x2 32x − 7x2 + 28x − x − 4 \| − 32x + 7 3x2 + 7x 4 − 5x2 3x3 − − 3x3 + 12x2 7x2 32x − 7x2 + 28x − 4x + 7 Multiply this term by divisor: 16 Change the signs and combine − + 4x 4(x − 4) = − 4x + 16 − − − 9 Remainder put over divisor and subtracted (due to negative) 207 3x2 + 7x 4 − − x 9 − 4 Example 260. Our Solution 6x3 − 8x2 + 10x + 103 2x + 4 Rewrite problem as long division 2x + 4 \| 6x3 − 8x2 + 10x + 103 Divide front terms: 6x3 2x = 3x2 2x + 4 \| − 3x2 6x3 − 6x3 − − 8x2 + 10x + 103 Multiply term by divisor: 3x2(2x + 4) = 6x3 + 12x2 12x2 Change the signs and combine 20x2 + 10x Bring down the next term 3x2 − 10x 6x3 2x + 4 \| 6x3 − − 8x2 + 10x + 103 Repeat, divide front terms: − 12x2 Multiply this term by divisor: 20x2 + 10x 40x − + 20x2 + 40x Change the signs and combine 10x(2x + 4) = − − 20x2 − − 20x2 2x 10x = − 3x2 − 2x + 4 \| − 6x3 6x3 50x + 103 Bring down the next term 10x + 25 − 8x2 + 10x + 103 Repeat, divide front terms: 12x2 − 20x2 + 10x − + 20x2 + 40x 50x 2x = 25 50x + 103 Multiply this term by divisor: 25(2x + 4) = 50x + 100 50x − − 100 Change the signs and combine 3 Remainder is put over divsor and added (
due to positive) 3x2 − 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, ﬁrst power, zero power (remember x0 = 1 so there is no variable on zero power). This is very important in long division, the variables must count down and no exponent can be skipped. If they don’t count down we must put them in order. If an exponent is skipped we will have to add a term to the problem, with zero for its coeﬃcient. This is demonstrated in the following example. 208 Example 261. 2x3 + 42 x + 3 − 4x Reorder dividend, need x2 term, add 0x2 for this 2x3 + 0x2 x + 3 \| − 4x + 42 Divide front terms: 2x3 x = 2x2 x + 3 − 2x2 2x3 + 0x2 − \| 6x2 2x3 6x2 − − − 4x + 42 Multiply this term by divisor: 2x2(x + 3) = 2x3 + 6x2 4x Change the signs and combine Bring down the next term 4x + 42 Repeat, divide front terms: − 6x2 x 6x = − Multiply this term by divisor: Change the signs and combine − 6x(x + 3) = 6x2 − − 18x 2x2 6x − − 2x3 + 0x2 6x2 2x3 6x2 4x + 6x2 + 18x − − − x + 3 \| − x + 3 \| − 14x + 42 Bring down the next term 4x + 42 Repeat, divide front terms: − 14x x = 14 2x2 6x + 14 − 2x3 + 0x2 6x2 2x3 6x2 − 4x − − + 6x2 + 18x 14x + 42 Multiply this term by divisor: 14(x + 3) = 14x + 42 14x − − 42 Change the signs and combine 0 No remainder 2x2 − 6x + 14 Our Solution It is important to take a moment to check each problem to verify that the exponents count down and no exponent is skipped. If so we will have to adjust the problem. Also,
this ﬁnal example illustrates, just as in regular long division, sometimes we have no remainder in a problem. World View Note: Paolo Ruﬃni was an Italian Mathematician of the early 19th century. In 1809 he was the ﬁrst to describe a process called synthetic division which could also be used to divide polynomials. 209 5.7 Practice - Divide Polynomials 10n Divide. 1) 20x4 + x3 + 2x2 4x3 3) 20n4 + n3 + 40n2 5) 12x4 + 24x3 + 3x2 7) 10n4 + 50n3 + 2n2 10n2 9) x2 2x − − x + 8 6x 71 11) n2 + 13n + 32 n + 5 13) v2 − v 15) a2 − 10 89 38 2v − 4a − − a − 8 17) 45p2 + 56p + 19 9p + 4 19) 10x2 − 10x 32x + 9 2 − 21) 4r2 1 r − − 4r + 3 23) n2 n 4 2 − − 25) 27b2 + 87b + 35 3b + 8 27) 4x2 33x + 28 5 − 4x − 29) a3 + 15a2 + 49a a + 7 35) x3 − 46x + 22 x + 7 37) 9p3 + 45p2 + 27p 9p + 9 5 − 39) r3 − r2 r 16r + 8 4 − − 41) 12n3 + 12n2 − 2n + 3 15n 4 − 43) 4v3 − 21v2 + 6v + 19 4v + 3 2) 5x4 + 45x3 + 4x2 9x 4) 3k3 + 4k2 + 2k 8k 6) 5p4 + 16p3 + 16p2 4p 8) 3m4 + 18m3 + 27m2 9m2 10) r2 12) b2 3r 53 − r − 9 − 10b + 16 − b 7 − 14) x2 + 4x − x + 7 26 16) x2 10x + 22 − x 4 − 18) 48k2 − 6k 70k + 16 2 − 20) n2 + 7n + 15 n + 4 22) 3m2
+ 9m 9 − 24) 2x2 3 3m − 5x − 2x + 3 8 − 26) 3v2 3v 32 9 − − 28) 4n2 38 23n − 4n + 5 − 36) 2n3 + 21n2 + 25n 2n + 3 38) 8m3 57m2 + 42 − 8m + 7 40) 2x3 + 12x2 + 4x 2x + 6 37 − 38b2 + 29b 4b 7 − 60 − 42) 24b3 − 210 55 − 30) 8k3 − 31) x3 41 − 26x x + 4 − 33) 3n3 + 9n2 64n − n + 6 68 − 32) x3 − 34) k3 − 66k2 + 12k + 37 8 k − 16x2 + 71x 56 − x 8 − 4k2 k 6k + 4 − 1 − Chapter 6 : Factoring 6.1 Greatest Common Factor.......................................................................212 6.2 Grouping................................................................................................216 6.3 Trinomials where a =1...........................................................................221 6.4 Trinomials where a 1.........................................................................226 6.5 Factoring Special Products....................................................................229 6.6 Factoring Strategy.................................................................................234 6.7 Solve by Factoring.................................................................................237 211 6.1 Factoring - Greatest Common Factor Objective: Find the greatest common factor of a polynomial and factor it out of the expression. The opposite of multiplying polynomials together is factoring polynomials. There are many beniﬁts of a polynomial being factored. We use factored polynomials to help us solve equations, learn behaviors of graphs, work with fractions and more. Because so many concepts in algebra depend on us being able to factor polynomials it is very important to have very strong factoring skills. In this lesson we will focus on factoring using the greatest common factor or GCF of a polynomial. When we multiplied polynomials, we multiplied monomials by polynomials by distributing, solving problems such as 4x2(2x2 − 12x3 + 32x. In this lesson we will work the same problem backwards. We will start with 8x2 12x3 + 32x and try and work backwards to the 4x2(2x 3x + 8) = 8x4
3x + 8). − − − To do this we have to be able to ﬁrst identify what is the GCF of a polynomial. We will ﬁrst introduce this by looking at ﬁnding the GCF of several numbers. To ﬁnd 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 quick mental math and it is shown in the following example Example 262. Find the GCF of 15, 24, and 27 15 3 = 5, 24 3 = 6, 27 3 GCF = 3 Our Solution = 9 Each of the numbers can be divided by 3 When there are variables in our problem we can ﬁrst ﬁnd the GCF of the num- 212 bers using mental math, then we take any variables that are in common with each term, using the lowest exponent. This is shown in the next example Example 263. GCF of 24x4y2z, 18x2y4, and 12x3yz5 24 6 = 4, 18 6 = 3, 12 6 = 2 Each number can be divided by 6 x2y x and y are in all 3, using lowest exponets GCF = 6x2y Our Solution To factor out a GCF from a polynomial we ﬁrst need to identify the GCF of all the terms, this is the part that goes in front of the parenthesis, then we divide each term by the GCF, the answer is what is left inside the parenthesis. This is shown in the following examples Example 264. 4x2 4 = x2, − 20x 4 4x2 20x + 16 GCF is 4, divide each term by 4 = − 4(x2 − 5x, − 16 4 = 4 This is what is left inside the parenthesis 5x + 4) Our Solution With factoring we can always check our solutions by multiplying (distributing in this case) out the answer and the solution should be the original equation. Example 265. 25x4 25x4 5x2 = 5x2, − 15x3 5x2 = − 3x, − 5x2(5x2 15x3 + 20x2 GCF is 5x2, divide each term by this 20x2 5x2 = 4 This is what is left inside
the parenthesis 3x + 4) Our Solution − Example 266. 3x3y2z + 5x4y3z5 − 4xy4 GCF is xy2, divide each term by this 213 3x3y2z xy2 = 3x2z, 5x4y3z5 xy2 = 5x3yz5, − 4xy4 xy2 = xy2(3x2z + 5x3yz5 − 4y2 This is what is left in parenthesis − 4y2) Our Solution World View Note: The ﬁrst recorded algorithm for ﬁnding the greatest common factor comes from Greek mathematician Euclid around the year 300 BC! Example 267. 21x3 7x = 3x2, 21x3 + 14x2 + 7x GCF is 7x, divide each term by this = 2x, 14x2 7x 7x(3x2 + 2x + 1) Our Solution 7x 7x = 1 This is what is left inside the parenthesis It is important to note in the previous example, that when the GCF was 7x and 7x was one of the terms, dividing gave an answer of 1. Students often try to factor out the 7x and get zero which is incorrect, factoring will never make terms dissapear. Anything divided by itself is 1, be sure to not forget to put the 1 into the solution. Often the second line is not shown in the work of factoring the GCF. We can simply identify the GCF and put it in front of the parenthesis as shown in the following two examples. Example 268. 12x5y2 − 2x3y2(6x2 6x4y4 + 8x3y5 GCF is 2x3y2, put this in front of parenthesis and divide 3xy2 + 4y3) Our Solution − Example 269. 18a4 b3 27a3b3 + 9a2b3 GCF is 9a2b3, divide each term by this − 9a2b3(2a2 3a + 1) Our Solution − Again, in the previous problem, when dividing 9a2b3 by itself, the answer is 1, not zero. Be very careful that each term is accounted for in your ﬁnal solution. 214 6.1 Practice - Greatest Common Factor Factor the common factor out
of each expression. 1) 9 + 8b2 3) 45x2 5) 56 − 25 − 35p 7) 7ab 35a2b − 3a2b + 6a3b2 9) − 5x2 11) − 13) 20x4 5x3 15x4 − − 30x + 30 − 15) 28m4 + 40m3 + 8 2) x 5 − 4) 1 + 2n2 80y 6) 50x − 8) 27x2y5 72x3y2 − 10) 8x3y2 + 4x3 12) − 32n9 + 32n6 + 40n5 14) 21p6 + 30p2 + 27 16) − 10x4 + 20x2 + 12x 18) 27y7 + 12y2x + 9y2 17) 30b9 + 5ab 15a2 − 56a3b 19) − 48a2b2 − 56a5b − 21) 20x8y2z2 + 15x5y2z + 35x3y3z 23) 50x2y + 10y2 + 70xz2 25) 30qpr 5qp + 5q − 18n5 + 3n3 21n + 3 − 27) 29) 31) − − − 40x11 − 20x12 + 50x13 50x14 − 32mn8 + 4m6n + 12mn4 + 16mn 20) 30m6 + 15mn2 25 22) 3p + 12q − − 15q2r2 24) 30y4z3x5 + 50y4z5 10y4z3x − 26) 28b + 14b2 + 35b3 + 7b5 28) 30a8 + 6a5 + 27a3 + 21a2 24x6 4x4 + 12x3 + 4x2 − 10y7 + 6y10 4y10x 8y8x − − 30) 32) − − 215 6.2 Factoring - Grouping Objective: Factor polynomials with four terms using grouping. The ﬁrst thing we will always do when factoring is try to factor out a GCF. This GCF is often a monomial like in the problem 5x y + 10xz the GCF is the monomial 5x, so we would have 5x(y + 2z). However
, 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 − − 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 common, factor it out − (2a + 5b)(3a 7b) Our Solution − In the same way we factored out a GCF of x we can factor out a GCF which is a binomial, (2a + 5b). This process can be extended to factor problems where there is no GCF to factor out, or after the GCF is factored out, there is more factoring that can be done. Here we will have to use another strategy to factor. We will use a process known as grouping. Grouping is how we will factor if there are four terms in the problem. Remember, factoring is like multiplying in reverse, so ﬁrst we will look at a multiplication problem and then try to reverse the process. Example 272. (2a + 3)(5b + 2) Distribute (2a + 3) into second parenthesis 5b(2a + 3) + 2(2a + 3) Distribute each monomial 10ab + 15b + 4a + 6 Our Solution The solution has four terms in it. We arrived at the solution by looking at the two parts, 5b(2a + 3) and 2(2a + 3). When we are factoring by grouping we will always divide the problem into two parts, the ﬁrst two terms and the last two terms. Then we can factor the GCF out of both the left and right sides. When we do this our hope is what is left in the parenthesis will match on both the left and right. If they match we can pull this matching GCF out front, putting the rest in parenthesis and we will be factored. The next example is the same problem worked backwards, factoring instead of multiplying. 216 Example 273. 10ab + 15b + 4a + 6 10ab + 15b + 4a + 6 5b(2a + 3) + 2(2a +
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 diﬀerence between the two we either have to do some adjusting or it can’t be factored using the grouping method. Consider the following example. Example 274. 6x2 + 9xy 6x2 + 9xy − 3x(2x + 3y) + 7( − − 14x 14x 2x − − − 21y 21y 3y) Split problem into two groups GCF on left is 3x, on right is 7 The signs in the parenthesis don′t match! when the signs don’t match on both terms we can easily make them match by factoring the opposite of the GCF on the right side. Instead of 7 we will use 7. This will change the signs inside the second parenthesis. − 3x(2x + 3y) 7(2x + 3y) (2x + 3y) is the same! Factor out this GCF − (2x + 3y)(3x 7) Our Solution − Often we can recognize early that we need to use the opposite of the GCF when factoring. If the ﬁrst term of the ﬁrst binomial is positive in the problem, we will also want the ﬁrst term of the second binomial to be positive. If it is negative then we will use the opposite of the GCF to be sure they match. Example 275. 8x − 8x − − 10y + 16 10y + 16 5xy 5xy x(5y − − Split the problem into two groups GCF on left is x, on right we need a negative, so we use (5y 8) is the same! Factor out this GCF − 2 8) (5y − − 2(5y 8)(x − − 8) − 2) Our Solution 217 Sometimes when factoring the GCF out of the left or right side there is no GCF to factor out. In this case we will use either the GCF of 1 or 1. Often this is all we need to be sure the
two binomials match. − Example 276. 12ab 12ab 2a(6b 14a − 14a Split the problem into two groups GCF on left is 2a, on right, no GCF, use (6b − 7) is the same! Factor out this GCF 6b + 7 6b + 7 7) − 1) Our Solution − − 1(6b 7)(2a 1 − 7) (6b − − − − − Example 277. 6x3 3x2(2x 5 6x3 15x2 + 2x − 15x2 + 2x 5) + 1(2x (2x − − − − 5 5) − − 5)(3x2 + 1) Our Solution − Split problem into two groups GCF on left is 3x2, on right, no GCF, use 1 (2x 5) is the same! Factor out this GCF Another problem that may come up with grouping is after factoring out the GCF on the left and right, the binomials don’t match, more than just the signs are different. In this case we may have to adjust the problem slightly. One way to do this is to change the order of the terms and try again. To do this we will move the second term to the end of the problem and see if that helps us use grouping. Example 278. 4a2 4a2 1(4a2 − − 14ab2 21b3 + 6ab − 21b3 + 6ab 14ab2 − 21b3) + 2ab(3 7b) 21b3 14ab2 14ab2 21b3 − 7b2(2a + 3b) − − − − − (2a + 3b)(2a − 4a2 + 6ab 4a2 + 6ab 2a(2a + 3b) Split the problem into two groups GCF on left is 1, on right is 2ab Binomials don′t match! Move second term to end Start over, split the problem into two groups GCF on left is 2a, on right is (2a + 3b) is the same! Factor out this GCF 7b2 − 7b2) Our Solution − When rearranging terms the problem can still be out of order. Sometimes after factoring out the GCF the terms are backwards. There are two ways that this can happen, one with addition
, one with subtraction. If it happens with addition, for 218 example the binomials are (a + b) and (b + a), we don’t 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 − 3xy 21x − 3x(y + 7) − − − (y + 7)(1 3x) Our Solution − Split the problem into two groups GCF on left is 1, on the right is y + 7 and 7 + y are the same, use either one 3x − However, if the binomial has subtraction, then we need to be a bit more careful. For example, if the binomials are (a a), we will factor out the opposite of the GCF on one part, usually the second. Notice what happens when we factor out b) and (b − − 1. − Example 280. − 1( − − 1(a a) Factor out (b b + a) Addition can be in either order, switch order − − 1 b) The order of the subtraction has been switched! − Generally we won’t show all the above steps, we will simply factor out the opposite of the GCF and switch the order of the subtraction to make it match the other binomial. Example 281. 8xy 8xy 12y + 15 − 12y 15 − 3) + 5(3 3) (2x − − − 5(2x 3)(4y 10x Split the problem into two groups GCF on left is 4y, on right, 5 10x Need to switch subtraction order, use 2x) Now 2x 3) 5) Our Solution − − − 4y(2x 4y(2y − − − − 3 match on both! Factor out this GCF 5 in middle − World View Note: Soﬁa Kovalevskaya of Russia was the ﬁrst woman on the editorial staﬀ of a mathematical journal in the late 19th century. She also did research on how the rings of Saturn rotated. 219 6.2 Practice - Grouping Factor each completely. 1) 40r3 8r2 − − 25r + 5 9n + 6 3) 3n3 2n2 − − 5) 15b
3 + 21b2 35b 49 − 7) 3x3 + 15x2 + 2x + 10 − 2) 35x3 10x2 − − 56x + 16 4) 14v3 + 10v2 7v 5 − − 6) 6x3 − 48x2 + 5x 40 − 8) 28p3 + 21p2 + 20p + 15 35 − 12) 42r3 − 9) 35x3 − 11) 7xy − 28x2 20x + 16 − 49x + 5y 13) 32xy + 40x2 + 12y + 15x 15) 16xy 17) 2xy − 56x + 2y − 8x2 + 7y3 7 − 28y2x 19) 40xy + 35x 21) 32uv − − 8y2 − 20u + 24v 7y 15 − − 10) 7n3 + 21n2 5n 15 − − 49r2 + 18r 14) 15ab 16) 3mn − − 6a + 5b3 − 8m + 15n 18) 5mn + 2m 20) 8xy + 56x 25n − − y 7 − − 21 − 2b2 − 40 10 22) 4uv + 14u2 + 12v + 42u 20x − 49a − 30y3 − 16b 23) 10xy + 30 + 25x + 12y 25) 3uv + 14u 27) 16xy 3x − − − 6u2 7v − 6x2 + 8y 24) 24xy + 25y2 26) 56ab + 14 − 220 6.3 Factoring - Trinomials where a = 1 Objective: Factor trinomials where the coeﬃcient of x2 is one. Factoring with three terms, or trinomials, is the most important type of factoring to be able to master. As factoring is multiplication backwards we will start with a multipication problem and look at how we can reverse the process. Example 282. x(x + 6) x2 + 6x (x + 6)(x − 4) Distribute (x + 6) through second parenthesis 4(x + 6) Distribute each monomial through parenthesis − 4x − x2 + 2x 24 Combine like terms 24 Our Solution − − You may notice that if you reverse the last three steps the process looks like grouping. This is because it is grouping! The GC
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 − Example 283. x2 + 6x x(x + 6) x2 + 2x − 4x − 4(x + 6) − − (x + 6)(x Split middle term into + 6x 24 24 Grouping: GCF on left is x, on right is 4x − − (x + 6) is the same, factor out this GCF 4 4) Our Solution − − 4x and not + 5x 3x? The reason is because 6x The trick to make these problems work is how we split the middle term. Why did we pick + 6x 4x is the only combination that works! So how do we know what is the one combination that works? To ﬁnd the correct way to split the middle term we will use what is called the ac method. In the next lesson we will discuss why it is called the ac method. The way the ac method works is we ﬁnd a pair of numers that multiply to a certain number and add to another number. Here we will try to multiply to get the last term and add to get the coeﬃcient of the middle term. In the previous − − 221 example that would mean we wanted to multiply to only numbers that can do this are 6 and This process is shown in the next few examples 4 (6 · − − − 4 = − 24 and add to + 2. The 4) = 2). 24 and 6 + ( − Example 284. x2 + 9x + 18 Want to multiply to 18, add to 9 x2 + 6x + 3x + 18 x(x + 6) + 3(x + 6) 6 and 3, split the middle term Factor by grouping (x + 6)(x + 3) Our Solution Example 285. x(x x2 − 3x − 1(x 3)(x 4 1, split the middle term 4x + 3 Want to multiply to 3, add to x + 3 3) 1) Our Solution − − Factor by grouping 3 and − x2 − 3) − (x − − − − Example 286. x2 x2 8x − 10x + 2x − 10) + 2
(x (x − − − − x(x − 20, add to 10 and 2, split the middle term 20 Want to multiply to 20 10) − Factor by grouping − 8 − 10)(x + 2) Our Solution Often when factoring we have two variables. These problems solve just like problems with one variable, using the coeﬃcients to decide how to split the middle term Example 287. a2 a(a − − a2 7ab 7b) (a − − − − 9ab + 14b2 Want to multiply to 14, add to 2ab + 14b2 7b) 2b(a 2b) Our Solution 7b)(a 9 2, split the middle term − − Factor by grouping 7 and − − − 222 As the past few examples illustrate, it is very important to be aware of negatives as we ﬁnd the pair of numbers we will use to split the middle term. Consier the following example, done incorrectly, ignoring negative signs Warning 288. x2 + 5x x2 + 2x + 3x x(x + 2) + 3(x 6 Want to multiply to 6, add 5 − 2 and 3, split the middle term 6 − Factor by grouping 2) −??? Binomials do not match! Because we did not use the negative sign with the six to ﬁnd our pair of numbers, the binomials did not match and grouping was not able to work at the end. Now the problem will be done correctly. Example 289. x2 + 6x x2 + 5x 6 Want to multiply to x 6 − 1(x + 6) 6 and Factor by grouping − − − − x(x + 6) 1, split the middle term 6, add to 5 − (x + 6)(x 1) Our Solution − You may have noticed a shortcut for factoring these problems. Once we identify the two numbers that are used to split the middle term, these are the two numbers in our factors! In the previous example, the numbers used to split the middle 1, our factors turned out to be (x + 6)(x term were 6 and 1). This pattern does not always work, so be careful getting in the habit of using it. We can use it however, when we have no number (technically we have a 1) in front of x2. In all the problems we have factored in this lesson there is no number in front of x2. If
this is the case then we can use this shortcut. This is shown in the next few examples. − − Example 290. x2 − 7x − 18 Want to multiply to 18, add to 9 and 2, write the factors − 7 − − 9)(x + 2) Our Solution (x − 223 Example 291. m2 − mn − 30n2 Want to multiply to 30, add to − 1 − 6, write the factors, don′t forget second variable 5 and 6n) Our Solution − (m + 5n)(m − It is possible to have a problem that does not factor. If there is no combination of numbers that multiplies and adds to the correct numbers, then we say we cannot factor the polynomial, or we say the polynomial is prime. This is shown in the following example. Example 292. x2 + 2x + 6 Want to multiply to 6, add to 2 1 3 Only possibilities to multiply to six, none add to 2 6 and 2 · Prime, can′t factor Our Solution · When factoring it is important not to forget about the GCF. If all the terms in a problem have a common factor we will want to ﬁrst factor out the GCF before we factor using any other method. Example 293. 3x2 3(x2 − − 24x + 45 GCF of all terms is 3, factor this out 8x + 15) Want to multiply to 15, add to 3, write the factors − 8 5 and 3) Our Solution − − 3(x − 5)(x − Again it is important to comment on the shortcut of jumping right to the factors, this only works if there is no coeﬃcient on x2. In the next lesson we will look at how this process changes slightly when we have a number in front of x2. Be careful not to use this shortcut on all factoring problems! World View Note: The ﬁrst person to use letters for unknown values was Francois Vieta in 1591 in France. He used vowels to represent variables we are solving for, just as codes used letters to represent an unknown message. 224 6.3 Practice - Trinomials where a = 1 Factor each completely. 1) p2 + 17p + 72 3) n2 5) x2 − − 9n + 8 9x 10 − 7) b2 + 12b + 32 9) x2 + 3x 70 − 8n
+ 15 11) n2 − 13) p2 + 15p + 54 15) n2 − 15n + 56 17) u2 8uv + 15v2 − 19) m2 + 2mn 8n2 − 11xy + 18y2 21) x2 − 23) x2 + xy − 25) x2 + 4xy 12y2 12y2 − 27) 5a2 + 60a + 100 29) 6a2 + 24a 192 − 31) 6x2 + 18xy + 12y2 33) 6x2 + 96xy + 378y2 2) x2 + x 4) x2 + x 72 30 − − 6) x2 + 13x + 40 8) b2 17b + 70 − 10) x2 + 3x 12) a2 6a − 14) p2 + 7p 18 27 30 − − − 16) m2 18) m2 − − 15mn + 50n2 3mn 40n2 − 20) x2 + 10xy + 16y2 22) u2 − 9uv + 14v2 24) x2 + 14xy + 45y2 26) 4x2 + 52x + 168 28) 5n2 − 45n + 40 30) 5v2 + 20v − 32) 5m2 + 30mn 25 90n2 162n2 − − 34) 6m2 36mn − 225 6.4 Factoring - Trinomials where a 1 Objective: Factor trinomials using the ac method when the coeﬃcient of x2 is not one. When factoring trinomials we used the ac method to split the middle term and then factor by grouping. The ac method gets it’s name from the general trinomial equation, a x2 + b x + c, where a, b, and c are the numbers in front of x2, x and the constant at the end respectively. World View Note: It was French philosopher Rene Descartes who ﬁrst used letters from the beginning of the alphabet to represent values we know (a, b, c) and letters from the end to represent letters we don’t know and are solving for (x, y, z). The ac method is named ac because we multiply a c to ﬁnd out what we want to multiply to. In the previous lesson we always multiplied to just c because there was no number in
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. · Example 294. 3x2 + 11x + 6 Multiply to ac or (3)(6) = 18, add to 11 3x2 + 9x + 2x + 6 The numbers are 9 and 2, split the middle term 3x(x + 3) + 2(x + 3) Factor by grouping (x + 3)(3x + 2) Our Solution When a = 1, or no coeﬃcient in front of x2, we were able to use a shortcut, using the numbers that split the middle term in the factors. The previous example illustrates an important point, the shortcut does not work when a 1. We must go through all the steps of grouping in order to factor the problem. Example 295. 15 Multiply to ac or (8)( 15 The numbers are 3) − Factor by grouping 15) = 120, add to 2 − − − 12 and 10, split the middle term 8x2 4x(2x 8x2 2x − 12x + 10x 3) + 5(2x − − − − − (2x − 3)(4x + 5) Our Solution 226 Example 296. 10x2 5x(2x 10x2 − 25x − 1(2x 5)(5x − 5) − (2x − − − − 27x + 5 Multiply to ac or (10)(5) = 50, add to 2x + 5 The numbers are 27 2, split the middle term 25 and − − Factor by grouping 5) 1) Our Solution − The same process works with two variables in the problem Example 297. 4x2 4x2 + 4xy 4x(x + y) xy − − 5xy − − 5y(x + y) − (x + y)(4x − 5y2 Multiply to ac or (4)( − 5y2 The numbers are 4 and 5) = 20, add to 5, split the middle term − − 1 Factor by grouping − 5y) Our Solution As always, when factoring we will ﬁrst look for a GCF before using any other method, including the ac method
. Factoring out the GCF ﬁrst also has the added bonus of making the numbers smaller so the ac method becomes easier. Example 298. 18x3 + 33x2 3x[6x2 + 11x 4x − − − − 2(2x + 5)] 30x GCF = 3x, factor this out ﬁrst 10] Multiply to ac or (6)( 10) = − 10] The numbers are 15 and Factor by grouping − 3x[6x2 + 15x 3x[3x(2x + 5) − 3x(2x + 5)(3x 2) Our Solution − 60, add to 11 4, split the middle term − As was the case with trinomials when a = 1, not all trinomials can be factored. If there is no combinations that multiply and add correctly then we can say the trinomial is prime and cannot be factored. Example 299. 3x2 + 2x 3(7) and 7 Multiply to ac or (3)( 7) = − 7(3) Only two ways to multiply to − 21, add to 2 21, it doesn′t add to 2 − − − Prime, cannot be factored Our Solution − 227 6.4 Practice - Trinomials where a 1 Factor each completely. 1) 7x2 48x + 36 − 3) 7b2 + 15b + 2 5) 5a2 − 7) 2x2 28 13a − 5x + 2 − 9) 2x2 + 19x + 35 11) 2b2 b 3 − − 13) 5k2 + 13k + 6 15) 3x2 − 17x + 20 17) 3x2 + 17xy + 10y2 19) 5x2 + 28xy 49y2 − 21) 6x2 − 23) 21k2 39x 21 − − 90 87k − 60x + 16 25) 14x2 − 27) 6x2 + 29x + 20 29) 4k2 17k + 4 − 31) 4x2 + 9xy + 2y2 33) 4m2 9mn − 35) 4x2 + 13xy + 3y2 − 9n2 2) 7n2 4) 7v2 6) 5n2 − − − 44n + 12 24v 16 − 4n 20 − 8) 3r2 4r
4 − − 10) 7x2 + 29x 30 − 26k + 24 12) 5k2 − 14) 3r2 + 16r + 21 16) 3u2 + 13uv 10v2 18) 7x2 2xy − − 20) 5u2 + 31uv − 5y2 28v2 − 22) 10a2 54a − 24) 21n2 + 45n 36 54 − − 26) 4r2 + r − 28) 6p2 + 11p 3 7 − 30) 4r2 + 3r 7 − 32) 4m2 + 6mn + 6n2 34) 4x2 − 36) 18u2 6xy + 30y2 3uv − 36v2 − 37) 12x2 + 62xy + 70y2 38) 16x2 + 60xy + 36y2 39) 24x2 − 52xy + 8y2 40) 12x2 + 50xy + 28y2 228 6.5 Factoring - Factoring Special Products Objective: Identify and factor special products including a diﬀerence of squares, perfect squares, and sum and diﬀerence of cubes. When factoring there are a few special products that, if we can recognize them, can help us factor polynomials. The ﬁrst is one we have seen before. When multiplying special products we found that a sum and a diﬀerence could multiply to a diﬀerence of squares. Here we will use this special product to help us factor Diﬀerence of Squares: a2 b2 = (a + b)(a b) − − If we are subtracting two perfect squares then it will always factor to the sum and diﬀerence of the square roots. Example 300. x2 (x + 4)(x − − Example 301. Subtracting two perfect squares, the square roots are x and 4 16 4) Our Solution 9a2 − (3a + 5b)(3a − 25b2 5b) Our Solution Subtracting two perfect squares, the square roots are 3a and 5b It is important to note, that a sum of squares will never factor. It is always prime. This can be seen if we try to use the ac method to factor x2 + 36. Example 302. x2 + 36 No bx term, we use 0x. x2 + 0x +
36 Multiply to 36, add to 0 12, 4 9, 6 36, 2 18, 3 1 · · · Prime, cannot factor Our Solution · · 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 diﬀerence of 4th powers. Because the square root of a fourth power is a square ( a4√ = a2), we can factor a diﬀerence of fourth powers just like we factor a diﬀerence of squares, to a sum and diﬀerence of the square roots. This will give us two factors, one which will be a prime sum of squares, and a second which will be a diﬀerence of squares which we can factor again. This is shown in the following examples. Example 303. a4 (a2 + b2)(a2 (a2 + b2)(a + b)(a b4 Diﬀerence of squares with roots a2 and b2 b2) The ﬁrst factor is prime, the second is a diﬀerence of squares! b) Our Solution − − − Example 304. x4 (x2 + 4)(x2 (x2 + 4)(x + 2)(x − − − 16 Diﬀerence of squares with roots x2 and 4 4) The ﬁrst factor is prime, the second is a diﬀerence of squares! 2) Our Solution Another factoring shortcut is the perfect square. We had a shortcut for multiplying a perfect square which can be reversed to help us factor a perfect square Perfect Square: a2 + 2ab + b2 = (a + b)2 A perfect square can be diﬃcult to recognize at ﬁrst glance, but if we use the ac method and get two of the same numbers we know we have a perfect square. Then we can just factor using the square roots of the ﬁrst and last terms and the sign from the middle. This is shown in the following examples. Example 305. x2 − (x − 6x + 9 Multiply to 9, add to The numbers are 3, the same! Perfect square 3)2 Use square roots from ﬁrst and last terms and
sign from the middle 6 − 3 and − − 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 ﬁrst and last terms and sign from the middle World View Note: The ﬁrst known record of work with polynomials comes from the Chinese around 200 BC. Problems would be written as “three sheafs of a good crop, two sheafs of a mediocre crop, and one sheaf of a bad crop sold for 29 dou. This would be the polynomial (trinomial) 3x + 2y + z = 29. Another factoring shortcut has cubes. With cubes we can either do a sum or a diﬀerence of cubes. Both sum and diﬀerence of cubes have very similar factoring formulas Sum of Cubes: a3 + b3 = (a + b)(a2 ab + b2) − Diﬀerence of Cubes: a3 b3 = (a − − b)(a2 + ab + b2) Comparing the formulas you may notice that the only diﬀerence is the signs in between the terms. One way to keep these two formulas straight is to think of SOAP. S stands for Same sign as the problem. If we have a sum of cubes, we add ﬁrst, a diﬀerence of cubes we subtract ﬁrst. O stands for Opposite sign. If we have a sum, then subtraction is the second sign, a diﬀerence would have addition for the second sign. Finally, AP stands for Always Positive. Both formulas end with addition. The following examples show factoring with cubes. Example 307. m3 (m 3)(m2 3m 9) Use formula, use SOAP to ﬁll in signs 27 We have cube roots m and 3 − (m − 3)(m2 + 3m + 9) Our Solution Example 308. 125p3 + 8r3 We have cube roots 5p and 2r (5p 2r)(25p2 10r 4r2) Use formula, use SOAP to ﬁll in signs (5p + 2r)(25p2 − 10r + 4r2) Our Solution The previous example illustrates an
important point. When we ﬁll in the trinomial’s ﬁrst 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 diﬀerence of cubes, students want to factor the second factor, the trinomial further. As a general rule, this factor will always be prime (unless there is a GCF which should have been factored out before using cubes rule). The following table sumarizes all of the shortcuts that we can use to factor special products Factoring Special Products Diﬀerence of Squares Sum of Squares Perfect Square Sum of Cubes Diﬀerence of Cubes − − b) b2 = (a + b)(a a2 a2 + b2 = Prime a2 + 2ab + b2 = (a + b)2 a3 + b3 = (a + b)(a2 a3 b3 = (a − ab + b2) b)(a2 + ab + b2) − − As always, when factoring special products it is important to check for a GCF ﬁrst. Only after checking for a GCF should we be using the special products. This is shown in the following examples Example 309. 72x2 2(36x2 2(6x + 1)(6x − − − 2 GCF is 2 1) Diﬀerence of Squares, square roots are 6x and 1 1) Our Solution Example 310. 24xy + 3y GCF is 3y 48x2y − 3y(16x2 − 8x + 1) Multiply to 16 add to 8 The numbers are 4 and 4, the same! Perfect Square 3y(4x − 1)2 Our Solution Example 311. 128a4b2 + 54ab5 GCF is 2ab2 2ab2(64a3 + 27b3) Sum of cubes! Cube roots are 4a and 3b 2ab2(4a + 3b)(16a2 − 12ab + 9b2) Our Solution 232 6.5 Practice - Factoring Special Products Factor each completely. 16 25 4 1) r2 3) v2 − − 5) p2 − 7) 9k2 − 9) 3x2 − 11) 16
x2 13) 18a2 4 27 36 − 50b2 − 2a + 1 15) a2 − 17) x2 + 6x + 9 6x + 9 19) x2 − 21) 25p2 10p + 1 − 23) 25a2 + 30ab + 9b2 25) 4a2 27) 8x2 20ab + 25b2 − 24xy + 18y2 − m3 64 29) 8 − 31) x3 − 33) 216 − 35) 125a3 u3 64 − 37) 64x3 + 27y3 39) 54x3 + 250y3 41) a4 43) 16 45) x4 81 z4 y4 − − − 47) m4 81b4 − 9 1 2) x2 − 4) x2 − 6) 4v2 − 8) 9a2 − 10) 5n2 1 1 20 − 12) 125x2 + 45y2 14) 4m2 + 64n2 16) k2 + 4k + 4 18) n2 − 8n + 16 20) k2 4k + 4 − 22) x2 + 2x + 1 24) x2 + 8xy + 16y2 26) 18m2 24mn + 8n2 − 28) 20x2 + 20xy + 5y2 30) x3 + 64 32) x3 + 8 34) 125x3 36) 64x3 216 − 27 − 38) 32m3 108n3 − 40) 375m3 + 648n3 42) x4 256 − 44) n4 − 46) 16a4 48) 81c4 1 − − b4 16d4 233 6.6 Factoring - Factoring Strategy Objective: Idenﬁty and use the correct method to factor various polynomials. With so many diﬀerent tools used to factor, it is easy to get lost as to which tool to use when. Here we will attempt to organize all the diﬀerent factoring types we have seen. A large part of deciding how to solve a problem is based on how many terms are in the problem. For all problem types we will always try to factor out the GCF ﬁrst. Factoring Strategy (GCF First!!!!!) 2 terms: sum or diﬀerence of squares or cubes: b2 = (a + b)(a a2 − b)
− a2 + b2 = Prime a3 + b3 = (a + b)(a2 ab + b2) − b3 = (a a3 − − 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 • • • We will use the above strategy to factor each of the following examples. Here the emphasis will be on which strategy to use rather than the steps used in that method. Example 312. 4x2 + 56xy + 196y2 GCF ﬁrst, 4 4(x2 + 14xy + 49y2) Three terms, try ac method, multiply to 49, add to 14 7 and 7, perfect square! 234 4(x + 7y)2 Our Solution Example 313. 5x 2y + 15xy 35x2 105x GCF ﬁrst, 5x − 5x(xy + 3y 5x[y(x + 3) − 7x 21) − 7(x + 3)] − − 5x(x + 3)(y Four terms, try grouping (x + 3) match! 7) Our Solution − Example 314. Example 315. 100x2 − 100(x2 100(x + 4)(x − − 400 GCF ﬁrst, 100 4) Two terms, diﬀerence of squares 4) Our Solution 108x3y2 39x2y2 + 3xy2 GCF ﬁrst, 3xy2 3xy2(36x2 3xy2[9x(4x − 3xy2(36x2 − 9x − 1(4x 1)(9x − 1) − 3xy2(4x − − − − 13x + 1) Thee terms, ac method, multiply to 36, add to 4x + 1) 1)] 1) Our Solution − − Factor by grouping 4, split middle term 9 and 13 − World View Note: Variables originated in ancient Greece where Aristotle would use a single capital letter to represent a number. Example 316. 5 + 625y3 GCF ﬁrst, 5 5(1 + 125y3) Two terms, sum of cubes 5(1 + 5y)(1 − 5y + 25y2) Our Solution It is important to
be comfortable and conﬁdent 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 − 2) 2x2 − 11x + 15 4) 16x2 + 48xy + 36y2 6) 20uv 60u3 5xv + 15xu2 − 8) 2x3 + 5x2y + 3y2x − 1) 24az 18ah + 60yz − 9uv + 4v2 − 2x3 + 128y3 3) 5u2 5) − 7) 5n3 + 7n2 6n − 9) 54u3 16 − 11) n2 n − 4xy + 3y2 13) x2 − 15) 9x2 − 17) m2 − 19) 36b2c 25y2 4n2 16xd − − 24b2d + 24xc 21) 128 + 54x3 23) 2x3 + 6x2y 20y2x − 25) n3 + 7n2 + 10n 27) 27x3 64 − 29) 5x2 + 2x 31) 3k3 33) mn − − 27k2 + 60k 12x + 3m 4xn − 35) 16x2 8xy + y2 − 37) 27m2 48n2 − 39) 9x3 + 21x2y 41) 2m2 + 6mn 60y2x 20n2 − − 128x3 22x 15 − 10) 54 − 12) 5x2 − 14) 45u2 16) x3 − 18) 12ab − 27y3 − 20) 3m3 6m2n − 22) 64m3 + 27n3 − 150uv + 125v2 18a + 6nb 9n − 24n2m 24) 3ac + 15ad2 + x2c + 5x2d2 26) 64m3 n3 − 28) 16a2 9b2 − 10x + 12 30) 2x2 − 32) 32x2 − 34) 2k2 + k 18y2 10 − 36) v2 + v 38) x3 + 4x2 40) 9n3 − 42) 2u2v2 3n2 11uv3 + 15v4 − 236 6.7 Factoring -
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 we did with the linear equations. One method that we can use to solve for the varaible is known as the zero product rule − Zero Product Rule: If ab = 0 then either a = 0 or b = 0 The zero product rule tells us that if two factors are multiplied together and the answer is zero, then one of the factors must be zero. We can use this to help us solve factored polynomials as in the following example. Example 317. 2x 3)(5x + 1) = 0 One factor must be zero − + 3 + 3 − (2x 3 = 0 or 5x + 1 = 0 1 − 1 − 5 1 or − 5 1 2x = 3 or 5x = 2 x = 3 2 − 2 5 Set each factor equal to zero Solve each equation Our Solution For the zero product rule to work we must have factors to set equal to zero. This means if the problem is not already factored we will factor it ﬁrst. Example 318. Factor using the ac method, multiply to 12, add to 1 3 and 4, split the middle term − 3 = 0 3 = 0 The numbers are 3) = 0 − Factor by grouping 4x2 4x2 + x 3x + 4x − 3) + 1(4x − (4x 3 = 0 or (4x 4x − 3)(x + 1) = 0 One factor must be zero Set each factor equal to zero 237 + 3 + 3 1 4x = 3 or x = − 4 4 Solve each equation 1 1 − − x = 3 4 or − 1 Our Solution Another important part of the zero product rule is that before we factor, the equation must equal zero. If it does not, we must move terms around so it does equal zero. Generally we like the x2 term to be positive. Example 319. x2 = 8x 15 − 8x + 15 8x + 15 = 0 − − 8x + 15 x2 Set equal to zero by moving terms to the left Factor using the ac method, multiply to 15, add to − (x 5)(
x − 5 = 0 or x x − + 5 + 5 3) = 0 The numbers are − 3 = 0 − + 3 + 3 3 − Set each factor equal to zero Solve each equation 5 and − 8 − x = 5 or x = 3 Our Solution Example 320. x2 − (x − 7x + 3x x2 4x 7)(x + 3) = 21 = − 21 = − + 9 + 9 12 = 0 − − − 4x − x2 9 Not equal to zero, multiply ﬁrst, use FOIL 9 Combine like terms 9 Move 9 to other side so equation equals zero − Factor using the ac method, mutiply to 12, add to 4 − − x − − − 6)(x + 2) = 0 The numbers are 6 and (x 6 = 0 or x + 2 = 0 2 2 Our Solution 2 Set each factor equal to zero Solve each equation 2 − x = 6 or − − + 6 + 6 − − Example 321. 3x2 + 4x 3x2 − 4x + 5 5 = 7x2 + 4x 3x2 − − − − 0 = 4x2 14 − 4x + 5 9 − Set equal to zero by moving terms to the right Factor using diﬀerence of squares 238 0 = (2x + 3)(2x − 2x + 3 = 0 or 2x 3 3 − 2x = or 2x = or 3) One factor must be zero Set each factor equal to zero Solve each equation Our Solution Most problems with x2 will have two unique solutions. However, it is possible to have only one solution as the next example illustrates. Example 322. 4x2 = 12x 9 − 12x + 9 − 4x2 12x + 9 − (2x − 12x + 9 = 0 3)2 = 0 2x − 3 = 0 − + 3 + 3 2x = 3 2 2 3 2 x = Set equal to zero by moving terms to left Factor using the ac method, multiply to 36, add to 6 and 6, a perfect square! − − Set this factor equal to zero Solve the equation 12 − Our Solution As always it will be important to factor out the GCF ﬁrst if we have one. This GCF is also a factor and must also be set equal to zero using the zero product rule. This may give us more than just two solution. The next few examples illustrate
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! − 4x2 4x(x 4x = 0 or x 4 4 Factor out the GCF of 4x − 8x = 0 − 2) = 0 One factor must be zero − 2 = 0 − + 2 + 2 Set each factor equal to zero Solve each equation x = 0 or 2 Our Solution 239 Example 324. 2x3 − 2x(x2 2x(x 14x2 + 24x = 0 7x + 12) = 0 − 3)(x − 3 = 0 or x 2x = 0 or or 3 or 4 Example 325. Factor out the GCF of 2x Factor with ac method, multiply to 12, add to 3 and 7 − 4) = 0 The numbers are − 4 = 0 − + 4 + 4 4 − Set each factor equal to zero Solve each equation Our Solutions − 6x2 + 21x 3(2x2 + 7x 2x − − − − 1(2x + 9)] = 0 3(2x2 + 9x Factor out the GCF of 3 Factor with ac method, multiply to 27 = 0 9) = 0 9) = 0 The numbers are 9 and 2 − Factor by grouping 3[x(2x + 9) − 3(2x + 9)(x 3 = 0 or 2x + 9 = 0 or x 3 0 1) = 0 One factor must be zero − 1 = 0 − + 1 + 1 9 or x = 1 − 2 Set each factor equal to zero Solve each equation − 9 9 − 2x = 2 18, add to 7 − x = 9 2 − or 1 Our Solution In the previous example, the GCF did not have a variable in it. When we set this factor equal to zero we got a false statement. No solutions come from this factor. Often a student will skip setting the GCF factor equal to zero if there is no variables in the GCF. Just as not all polynomials cannot factor, all equations cannot be solved by factoring. If an equation does not factor we will have to solve it using another method. These other methods are saved for another section. World View Note: While factoring works great to solve problems with x2, Tartaglia, in 16th century Italy, developed a method to solve problems
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 Cardan’s 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 is possible, there are a few properties of square roots that we have not covered yet and thus it is common to break a rule of roots that we are not aware of at this point. The short reason we want to avoid this for now is because taking a square root will only give us one of the two answers. When we talk about roots we will come back to problems like these and see how we can solve using square roots in a method called completing the square. For now, never take the square root of both sides! 240 6.7 Practice - Solve by Factoring Solve each equation by factoring. 1) (k − 7)(k + 2) = 0 1)(x + 4) = 0 3) (x − 5) 6x2 150 = 0 − 7) 2n2 + 10n 28 = 0 − 9) 7x2 + 26x + 15 = 0 11) 5n2 13) x2 − 4x − 9n 2 = 0 − 8 = − 15) x2 5x 1 = − 17) 49p2 + 371p − 8 5 − − 163 = 5 − 19) 7x2 + 17x 20 = 8 − − 21) 7r2 + 84 = 49r − 6x = 16 23) x2 − 25) 3v2 + 7v = 40 2) (a + 4)(a − 4) (2x + 5)(x 6) p2 + 4p − 3) = 0 7) = 0 − 32 = 0 8) m2 − 10) 40r2 12) 2b2 m 30 = 0 − 285r − 3b − 2 = 0 − − 280 = 0 14) v2 16) a2 − − 8v 3 = − 6a + 6 = 3 2 − − 18) 7k2 + 57k + 13 = 5 20) 4n2 − 13n + 8 = 5 22) 7m2 − 224 = 28m 24) 7n2 28n = 0 − 26) 6b2 = 5 + 7b 28) 9n2 + 39n = 36 30) a2
+ 7a − − 9 = − 3 + 6a 27) 35x2 + 120x = 45 − 23 = 6k 29) 4k2 + 18k − 7 − 32) x2 + 10x + 30 = 6 31) 9x2 − 46 + 7x = 7x + 8x2 + 3 33) 2m2 + 19m + 40 = 2m − 35) 40p2 + 183p 168 = p + 5p2 − 34) 5n2 + 41n + 40 = 2 − 80 = 3x 36) 24x2 + 11x − 241 Chapter 7 : Rational Expressions 7.1 Reduce Rational Expressions..................................................................243 7.2 Multiply and Divide................................................................................248 7.3 Least Common Denominator..................................................................253 7.4 Add and Subtract...................................................................................257 7.5 Complex Fractions..................................................................................262 7.6 Proportions.............................................................................................268 7.7 Solving Rational Equations....................................................................274 7.8 Application: Dimensional Analysis.........................................................279 242 7.1 Rational Expressions - Reduce Rational Expressions Objective: Reduce rational expressions by dividing out common factors. Rational expressions are expressions written as a quotient of polynomials. Examples of rational expressions include: x2 x2 − − x 12 − 9x + 20 and 3 − 2 x and a b b a − − and 3 2 As rational expressions are a special type of fraction, it is important to remember with fractions we cannot have zero in the denominator of a fraction. For this reason, rational expressions may have one more excluded values, or values that the variable cannot be or the expression would be undeﬁned. Example 326. State the excluded value(s): Denominator cannot be zero Factor Set each factor not equal to zero Subtract 5 from second equation x2 1 − 3x2 + 5x 3x2 + 5x 0 x(3x + 5) 0 x 0 or 3x + 5 0 5 5 − − 3x − 5 Divide by or − 3 Our Solution Second equation is solved This means we can use any value for x in the equation except for 0 and − 5 3. We 243 can however, evaluate any other value in the expression. World View Note: The number zero was not widely accepted in mathematical thought around the world for many years. It was the Mayans of Central America who ﬁrst used zero
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 − x2 + 6x + 8 when x = 6 − Substitute − 5 in for each variable 6)2 ( − 6)2 + 6( ( − 36 36 + 6( 4 6) + 8 4 6) + 8 − − − − 4 − 36 + 8 36 − 36 Exponents ﬁrst Multiply Add and subtract 32 8 Reduce 4 Our Solution Just as we reduced the previous example, often a rational expression can be reduced, even without knowing the value of the variable. When we reduce we divide out common factors. We have already seen this with monomials when we discussed properties of exponents. If the problem only has monomials we can reduce the coeﬃcients, and subtract exponents on the variables. Example 328. 15x4y2 25x2y6 Reduce, subtract exponents. Negative exponents move to denominator 3x2 5y4 Our Solution 244 However, if there is more than just one term in either the numerator or denominator, we can’t divide out common factors unless we ﬁrst factor the numerator and denominator. Example 329. Example 330. 16 28 8x2 − 28 8(x2 2) − 7 2(x2 − 2) Denominator has a common factor of 8 Reduce by dividing 24 and 8 by 4 Our Solution 9x 18x 3 6 − − Numerator has a common factor of 3, denominator of 6 3(3x 6(3x 1) 1) − − Divide out common factor (3x 1) and divide 3 and 6 by 3 − 1 2 Our Solution Example 331. x2 25 − x2 + 8x + 15 Numerator is diﬀerence of squares, denominator is factored using ac (x + 5)(x 5) (x + 3)(x + 5) − Divide out common factor (x + 5) x 5 − x + 3 Our Solution It is important to remember we cannot reduce terms, only factors. This means if between the parts we want to reduce we cannot. In the prethere are any + or − vious example we had the solution x x + 3, we cannot divide out the x’s because ) not factors (separ
ated by multiplication). they are terms (separated by + or − 5 − 245 7.1 Practice - Reduce Rational Expressions Evaluate 1) 4v + 2 6 when v = 4 3) 5) x 3 − x2 4x + 3 when x = − b + 2 − b2 + 4b + 4 when b = 0 4 4) 2) b 3b 3 9 when b = − − a + 2 2 − a2 + 3a + 2 when a = 1 − 3 when n = 4 − 6 6) n2 − n n − State the excluded values for each. 7) 3k2 + 30k k + 10 9) 15n2 10n + 25 11) 10m2 + 8m 10m 13) r2 + 3r + 2 5r + 10 15) b2 + 12b + 32 b2 + 4b 32 − Simplify each expression. 17) 21x2 18x 19) 24a 40a2 21) 32x3 8x4 23) 18m − 60 24 25) 20 4p + 2 27) x + 1 x2 + 8x + 7 29) 32x2 28x2 + 28x 31) n2 + 4n n2 12 − 7n + 10 33) v2 − 9v + 54 4v − 35) 12x2 30x2 42x 42x − − 10 − 10a + 4 37) 6a 39) 2n2 + 19n − 9n + 90 10 8) 10) 27p 18p2 36p − x + 10 8x2 + 80x 12) 10x + 16 6x + 20 14) 6n2 21n − 6n2 + 3n 16) 10v2 + 30v 35v2 5v − 18) 12n 4n2 20) 21k 24k2 22) 90x2 20x 24) 10 81n3 + 36n2 − − 9 81 26) n 9n 28) 28m + 12 36 30) 49r + 56 56r 32) b2 + 14b + 48 b2 + 15b + 56 34) 30x 90 − 50x + 40 12k + 32 64 − k2 − 38) 9p + 18 p2 + 4p + 4 40) 3x2 5x2 29x + 40 30x 80 − − − 246 − 60 36) k2 41) 8m + 16 20m 12 − 42
) 56x 48 − 24x2 + 56x + 32 43) 2x2 3x2 10x + 8 7x + 4 − − 45) 7n2 32n + 16 16 − 4n − 47) n2 2n + 1 − 6n + 6 49) 7a2 6a2 26a 45 − 34a + 20 − − 44) 50b 80 − 50b + 20 46) 35v + 35 21v + 7 48) 56x 48 − 24x2 + 56x + 32 50) 4k3 9k3 2k2 2k − 18k2 + 9k − − 247 7.2 Rational Expressions - Multiply & Divide Objective: Multiply and divide rational expressions. Multiplying and dividing rational expressions is very similar to the process we use to multiply and divide fractions. Example 332. 15 49 · 14 45 1 7 · 2 3 2 21 First reduce common factors from numerator and denominator (15 and 7) Multiply numerators across and denominators across Our Solution The process is identical for division with the extra ﬁrst step of multiplying by the reciprocal. When multiplying with rational expressions we follow the same process, ﬁrst divide out common factors, then multiply straight across. 248 Example 333. 25x2 9y8 · 24y4 55x7 5 3y4 · 8 11x5 40 33x5y4 Reduce coeﬃcients by dividing out common factors (3 and 5) Reduce, subtracting exponents, negative exponents in denominator Multiply across Our Solution Division is identical in process with the extra ﬁrst step of multiplying by the reciprocal. Example 334. a4b2 a ÷ b4 4 a4b2 a · 4 b4 Multiply by the reciprocal Subtract exponents on variables, negative exponents in denominator a3 1 · 4 b2 Multiply across 4a3 b2 Our Solution Just as with reducing rational expressions, before we reduce a multiplication problem, it must be factored ﬁrst. Example 335. x2 − x2 + x 9 x2 8x + 16 − 3x + 9 20 · Factor each numerator and denominator − (x + 3)(x (x 3) − 4)(x + 5) · − (x 4)(x − − 3(x + 3) 4 Divide out common factors (x + 3
) and (x 4) − Multiply across 249 (x 3)(x − − 3(x + 5) 4) Our Solution Again we follow the same pattern with division with the extra ﬁrst step of multiplying by the reciprocal. Example 336. 12 8 ÷ x − 2x − 5x2 + 15x x2 + x 2 − Multiply by the reciprocal x2 + x 2 5x2 + 15x − Factor each numerator and denominator x2 x2 − − x2 x2 12 8 · x − 2x − − − 4)(x + 3) (x − (x + 2)(x 4) · − (x + 2)(x − 5x(x + 3) 1) Divide out common factors: (x 4) and (x + 3) and (x + 2) − Multiply across Our Solution 1 1 · 1 1 x x − 5x − 5x We can combine multiplying and dividing of fractions into one problem as shown below. To solve we still need to factor, and we use the reciprocal of the divided fraction. Example 337. a2 + 7a + 10 a2 + 6a + 5 · a + 1 a2 + 4a + 4 ÷ a 1 − a + 2 Factor each expression (a + 5)(a + 2) (a + 5)(a + 1) · (a + 1) (a + 2)(a + 2) ÷ (a 1) − (a + 2) Reciprocal of last fraction (a + 5)(a + 2) (a + 5)(a + 1) · (a + 1) (a + 2)(a + 2) · (a + 2) (a 1) − 1 − 1 a Divide out common factors (a + 2), (a + 2), (a + 1), (a + 5) Our Solution World View Note: Indian mathematician Aryabhata, in the 6th century, published a work which included the rational expression n(n + 1)(n + 2) for the sum of the ﬁrst n squares (11 + 22 + 32 +. + n2) 6 250 7.2 Practice - Multiply and Divide Simplify each expression. 1) 8x2 9 · 9 2 7 5n 3) 9n 2n · 5) 5x2 4 · 7) 7 (m 6) − m 6 − 6 5 · 5
m(7m 7(7m 5) 5) − − 9) 7r 7r(r + 10) ÷ r (r 6 6)2 − − 11) 25n + 25 5 4 30n + 30 · 13) x 15) x2 − 10 35x + 21 ÷ 6x − x + 5 − 7 7 35x + 21 x + 5 x 7 − · 17) 8k 24k2 40k ÷ − 15k 1 − 25 19) (n 8) − 21) 4m + 36 m + 9 · 3x 6 12x 23) − − 80 6 − 10n · m 5 − 5m2 24(x + 3) 12 6n − 13n + 42 n2 − − 27) n 6n 7 12 · 29) 27a + 36 9a + 63 ÷ 12x + 32 6x 16 · − − − 31) x2 x2 − 6a + 8 2 7x2 + 14x 7x2 + 21x 33) (10m2 + 100m) 35) 7p2 + 25p + 12 6p + 48 18m3 20m2 36m2 40m − − · 3p 21p2 · − 8 − 44p 32 − 2) 8x 3x ÷ 4) 9m 5m2 · 4 7 7 2 8 10 6) 10p 8) 5 ÷ 7 10(n + 3) ÷ 10) 6x(x + 4) x 3 · n 2 (n + 3)(n − 2) − (x 3)(x − 6x(x 6) − 6) 5 12 − − − b − 9 b 12) b2 14) v − 1 − 4 b − − b2 12 ÷ 4 11v + 10 v2 · − 8a + 80 8 6 · p − 12p + 32 ÷ 8 − 16) 1 − a 18) p2 x2 · 2r 7r + 42 20) x2 − x 7x + 10 2 − 22) 2r r + 6 ÷ 12n − n + 7 24) 2n2 1 10 p − x + 10 x − − 20 54 − ÷ (2n + 6) 35v v 20 − 9 − ÷ 6x3 + 6x2 x2 + 5x 24 − · 7k2 8k2 28) x2 + 11x + 24 6x3 + 18
x2 k 7 − k 30) k2 − − 12 · 32) 9x3 + 54x2 x2 + 5x 28k 56k − − x2 + 5x − 10x2 14 14 · − 7 9n + 54 10n + 50 34) n − 2n n2 − 36) 7x2 − 49x2 + 7x − 35 ÷ 66x + 80 72 ÷ − 7x2 + 39x 2 49x + 7x − − 70 72 15 25) b + 2 40b2 − 24b(5b 3) − 26) 21v2 + 16v 3v + 4 − 16 37) 10b2 30b + 20 · 30b + 20 2b2 + 10b 39) 7r2 24 53r − − 7r + 2 ÷ 49r + 21 49r + 14 38) 35n2 49n2 12n 32 − 91n + 40 · − − 7n2 + 16n 5n + 4 − 40) 12x + 24 10x2 + 34x + 28 · 15x + 21 5 251 41) x2 2x 1 4 · − − x2 x2 − − x 4 2 ÷ − 2 x2 + x 3x − − 6 42) a3 + b3 a2 + 3ab + 2b2 · 3a2 6b 3a − 3ab + 3b2 ÷ − a2 4b2 − a + 2b 43) x2 + 3x + 9 x2 + x 12 · − x2 + 2x x3 − 8 − 27 ÷ x2 − x2 4 − 6x + 9 44) x2 + 3x 10 − x2 + 6x + 5 · 2x2 x − 2x2 + x 3 6 ÷ − − 8x + 20 6x + 15 252 7.3 Rational Expressions - Least Common Denominators Objective: denominators to match this common denominator. Idenﬁty the least common denominator and build up As with fractions, the least common denominator or LCD is very important to working with rational expressions. The process we use to ﬁnd the LCD is based on the process used to ﬁnd the LCD of intergers. Example 338. Find the LCD of 8 and 6 Consider multiples of the larger number 8, 16, 24. 24 is the ﬁrst multiple of 8
that is also divisible by 6 24 Our Solution When ﬁnding the LCD of several monomials we ﬁrst ﬁnd the LCD of the coeﬃcients, then use all variables and attach the highest exponent on each variable. Example 339. Find the LCD of 4x2y5 and 6x4y3z6 First ﬁnd the LCD of coeﬃcients 4 and 6 12 is the LCD of 4 and 6 12 x4y5z6 Use all variables with highest exponents on each variable 12x4y5z6 Our Solution The same pattern can be used on polynomials that have more than one term. However, we must ﬁrst factor each polynomial so we can identify all the factors to be used (attaching highest exponent if necessary). Example 340. Find the LCD of x2 + 2x (x − 3 and x2 x 12 − − 4)(x + 3) Factor each polynomial LCD uses all unique factors − 1)(x + 3) and (x (x − 1)(x + 3)(x − 4) Our Solution − Notice we only used (x + 3) once in our LCD. This is because it only appears as a factor once in either polynomial. The only time we need to repeat a factor or use an exponent on a factor is if there are exponents when one of the polynomials is factored 253 Example 341. Find the LCD of x2 10x + 25 and x2 14x + 45 − − (x − 5)2 and (x (x 5)(x 5)2(x − − Factor each polynomial LCD uses all unique factors with highest exponent 9) 9) Our Solution − − 5)(x The previous example could have also been done with factoring the ﬁrst polynomial to (x 5) twice in the LCD because it showed up twice in one of the polynomials. However, it is the author’s suggestion to use the exponents in factored form so as to use the same pattern (highest exponent) as used with monomials. 5). Then we would have used (x − − − Once we know the LCD, our goal will be to build up fractions so they have matching denominators. In this lesson we will not be adding and subtracting fractions, just building them up to a common denominator. We
can build up a fraction’s denominator by multipliplying the numerator and denoinator by any factors that are not already in the denominator. Example 342. 5a 3a2b =? 6a5b3 Idenﬁty what factors we need to match denominators 2a3b2 3 · 2 = 6 and we need three more a′s and two more b′s 5a 3a2b 2a3b2 2a3b2 Multiply numerator and denominator by this 10a4b2 6a5b3 Our Solution Example 343. x 2 − x + 4 =? x2 + 7x + 12 (x + 4)(x + 3) Factor to idenﬁty factors we need to match denominators (x + 3) The missing factor x2 + x 6 − (x + 4)(x + 3) Multiply numerator and denominator by missing factor, FOIL numerator Our Solution 254 As the above example illustrates, we will multiply out our numerators, but keep our denominators factored. The reason for this is to add and subtract fractions we will want to be able to combine like terms in the numerator, then when we reduce at the end we will want our denominators factored. Once we know how to ﬁnd the LCD and how to build up fractions to a desired denominator we can combine them together by ﬁnding a common denominator and building up those fractions. Example 344. Build up each fraction so they have a common denominator 5a 4b3c and 3c 6a2b First identify LCD 12a2b3c Determine what factors each fraction is missing First: 3a2 Second: 2b2c Multiply each fraction by missing factors 5a 4b3c 3a2 3a2 and 3c 6a2b 2b2c 2b2c 15a3 12a2b3c and 6b2c2 12a2b3c Our Solution Example 345. Build up each fraction so they have a common denominator 5x 5x 6 − 6)(x + 1) − x2 (x − and x 2 − x2 + 4x + 3 Factor to ﬁnd LCD (x + 1)(x + 3) Use factors to ﬁnd LCD LCD: (x 6)(x + 1)(x + 3) −
First: (x + 3) Second: (x Identify which factors are missing 6) Multiply fractions by missing factors − 5x 6)(x + 1) x + 3 x + 3 (x − and x 2 − (x + 1)(x + 3) x x 6 6 − − Multiply numerators 5x2 + 15x 6)(x + 1)(x + 3) and (x (x − x2 8x + 12 6)(x + 1)(x + 3) − − Our Solution World View Note: When the Egyptians began working with fractions, they expressed all fractions as a sum of unit fraction. Rather than 4 5, they would write 4 + 1 2 + 1 the fraction as the sum, 1 20. An interesting problem with this system is 3 + 1 5 is also equal to the sum 1 this is not a unique solution, 4 6 + 1 10. 5 + 1 255 7.3 Practice - Least Common Denominator Build up denominators. 1) 3 8 =? 48 3) a x =? xy 5) 7) 2 3a3b2c =? 9a5b2c4 2 x + 4 =? x2 − 16 9) x 4 x + 2 = −? x2 + 5x + 6 Find Least Common Denominators 11) 2a3, 6a4b2, 4a3b5 13) x2 3x, x 3, x − − 15) x + 2, x 4 − 25, x + 5 17) x2 − 2) a 5 =? 5a 4) 6) 5 2x2 =? 8x3y 4 3a5b2c4 =? 9a5b2c4 8) x + 1 x − 10) x 3 =? 6x + 9 x2 − 6 x + 3 = −? 2x x2 − 15 − 12) 5x2y, 25x3y5z 14) 4x − 16) x, x 8, x 2, 4 − 7, x + 1 18) x2 6x + 9 − 9, x2 − − 7x + 10, x2 19) x2 + 3x + 2, x2 + 5x + 6 20) x2 − Find LCD and build up each fraction 2x − − 15, x2 + x 6 − 21) 3a 5b2, 2 10a3b 23 25
) x2 16 − 27) x + 1 x2 36 −,, 3x 8x + 16 x2 − 2x + 3 x2 + 12x + 36 29) 4x x − x2 3x x − 5 − x2, 2 x, − x − 3 6 6x 5x + 1 3x − −, 10 x 4 − 5 3x + 1 x − − 12 3x 6x + 8 −,, 2x x2 + 4x + 3 x − x2 + x 2 −, 5 x2 + 3x 20 10 − x2 x2 x2 22) 24) 26) 28) 30) 256 7.4 Rational Expressions - Add & Subtract Objective: Add and subtract rational expressions with and without common denominators. Adding and subtracting rational expressions is identical to adding and subtracting with integers. Recall that when adding with a common denominator we add the numerators and keep the denominator. This is the same process used with rational expressions. Remember to reduce, if possible, your ﬁnal answer. Example 346. x − 4 − 2x − x2 + 8 x2 x2 x + 8 2x − − 2x + 4 2x − − Same denominator, add numerators, combine like terms Factor numerator and denominator 8 8 2(x + 2) (x + 2)(x 4) − 2 − 4 x Divide out (x + 2) Our Solution 257 Subtraction with common denominator follows the same pattern, though the subtraction can cause problems if we are not careful with it. To avoid sign errors we will ﬁrst distribute the subtraction through the numerator. Then we can treat it like an addition problem. This process is the same as “add the opposite” we saw when subtracting with negatives. Example 347. 6x 3x 12 6 − − − 12 6 6x 3x − − + − 15x 3x 6 − 6 − 15x + 6 3x 6 − 9x − 3x 6 − 6 − − 3(3x + 2) 3(x 2) − Add the opposite of the second fraction (distribute negative) Add numerators, combine like terms Factor numerator and denominator Divide out common factor of 3 − (3x + 2) x 2 − Our Solution World View Note: The Rhind papyrus of Egypt from 1650 BC gives some of the earliest known symbols for
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 don’t have a common denominator we will have to ﬁnd the least common denominator (LCD) and build up each fraction so the denominators match. The following example shows this process with integers. Example 348 10 12 3 3 3 12 + The LCD is 12. Build up, multiply 6 by 2 and 4 by 3 Multiply Add numerators 258 13 12 Our Solution The same process is used with variables. Example 349. 7a 3a2b + 4b 6ab4 The LCD is 6a2b4. We will then build up each fraction 2b3 2b3 7a 3a2b + 4b 6ab4 a a Multiply ﬁrst fraction by 2b3 and second by a 4ab 6a2b4 Add numerators, no like terms to combine 14ab3 6a2b4 + 14ab3 + 4ab 6a2b4 2ab(7b3 + 2) 6a2b4 Factor numerator Reduce, dividing out factors 2, a, and b 7b3 + 2 3ab3 Our Solution The same process can be used for subtraction, we will simply add the ﬁrst step of adding the opposite. Example 350. 4 5a − 7b 4a2 Add the opposite 4 5a + − 7b 4a2 LCD is 20a2. Build up denominators 4a 4a 4 5a + − 7b 4a2 5 5 Multiply ﬁrst fraction by 4a, second by 5 16a 35b − 20a2 Our Solution If our denominators have more than one term in them we will need to factor ﬁrst to ﬁnd the LCD. Then we build up each denominator using the factors that are 259 missing on each fraction. Example 351. 3a 6 8a + 4 8 4(2a + 1) 8 + Factor denominators to ﬁnd LCD LCD is 8(2a + 1), build up each fraction 2 2 6 4(2a + 1) + 3a 8 2a + 1 2a + 1 Multiply ﬁrst fraction by 2, second by 2a + 1 12 8(2a + 1) + 6a2 +
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 − x2 Add the opposite (distribute negative) 4 − + − x2 − 4)(x x 1 − 7x + 12 3) (x − − Factor denominators to ﬁnd LCD LCD is (x 4)(x − − 3), build up each fraction − + − x2 − x 1 − 7x + 12 Only ﬁrst fraction needs to be multiplied by x 3 − x2 (x 2x − 3)(x 3 4) − − + (x x − 3)(x − − 1 − 4) Add numerators, combine like terms − (x (x (x x2 3x − 3)(x 4 4) − − − 4)(x + 1) 3)(x 4) − − − x + 1 x 3 − Factor numerator Divide out x 4 factor − Our Solution 260 7.4 Practice - Add and Subtract Add or subtract the rational expressions. Simplify your answers 2 a + 3 whenever possible. a + 3 + 4 1 + 2t − 2x2 + 3 7 − 1 − t t 3) t2 + 4t 1) 5) x2 x2 − − 5x + 9 6x + 5 x2 6x + 5 − − 5 8r 7) 5 6r − 9t3 + 5 6t2 9) 8 11) a + 2 13) x 2 − 1 − 4x − 15) 5x + 3y a 4 − 4 2x + 3 x 3x + 4y xy2 2x2y − 2z 3z z + 1 1 − z − 8 4 − x2 − t 3 x + 2 5 12 4 3x + 3 t 4t − − 3 − 2 5x2 + 5x − x2 + 5x + 6 − x x2 + 15x + 56 − 18 17) 19) 21) 23) 25) 27) 29) 31) 33) 35) 6 − − x2 9 − 4 x2 + 2x 3 − 5x x x2 − 2x x2 x2 1 − − x + 1 2x 35 + x + 6 x2 + 7x + 10 37) 4 − a2 − 2z 39) − 2 a a 3 −
− − a2 9 − 2z + 3z 2z + 1 − x2 + 3x + 2 + 3x − 2x − 3 1 41) 1 − x2 + 5x + 6 2 x2 + 3x + 2 7 x2 + 13x + 42 3 4z2 − 1 38) 40) 43) x2 2x + 7 2x − 3 − − 3x 2 − x2 + 6x + 5 44) 261 4 a2 + 5a 6 − 2) 4) 6) 3 8) x2 x 2 − − a2 + 3a 6x x 8 − 2 − 7 − x2 6 − a2 + 5a x + 4 xy2 + 3 8 + x 3a2 + 5a + 1 x2y − 12 9a − 1 3 10) x + 5 12) 2a 14) 2c x − 2 c + d cd2 x + 1 d − c2d − 1 + 2 2 5 + 3 25 + x x + 3 + 4 − 4x − 2 x2 4x x x + 5 (x + 3)2 4a 3a − x 6a 20 + 9a 5 + x − x − 5 30 26) x 16) 18) 20) 22) 24) 28) 30) 32) 34) − 2x − 2x − x2 x2 1 − 9 + 4x 2x 3 x2 + 5x + 4 5 x2 + x 6 − 3 5x + 6 x2 x2 − − − x − 1 3 − x2 + 3x + 2 + x + 5 3x + 6 + x x2 4 x2 + 4x + 3 − 4y − 2r y2 r2 2 y + 1 2 y − 1 − s2 + 1 r + s − 4x + 3 + 4x + 5 x2 + 4x − 1 r s − x + 2 8 x2 + 6x + 8 + 2x − x2 + 3x + 2 − 5 − 3 36) 3x + 2 42) x2 − 3x 7.5 Rational Expressions - Complex Fractions Objective: Simplify complex fractions by multiplying each term by the least common denominator. Complex fractions have fractions in either the numerator, or denominator, or usually both. These fractions can be simpliﬁed in one of two ways. This will be illustrated ﬁrst with
integers, then we will consider how the process can be expanded to include expressions with variables. The ﬁrst method uses order of operations to simplify the numerator and denominator ﬁrst, then divide the two resulting fractions by multiplying by the reciprocal. Example 353 12 8 12 − 5 6 + 3 6 5 12 4 3 3 4 1 4 5 16 5 12 5 4 Get common denominator in top and bottom fractions Add and subtract fractions, reducing solutions To divide fractions we multiply by the reciprocal Reduce Multiply Our Solution The process above works just ﬁne to simplify, but between getting common denominators, taking reciprocals, and reducing, it can be a very involved process. Generally we prefer a diﬀerent method, to multiply the numerator and denominator of the large fraction (in eﬀect each term in the complex fraction) by the least common denominator (LCD). This will allow us to reduce and clear the small fractions. We will simplify the same problem using this second method. Example 354 LCD is 12, multiply each term 262 2(12) 1(12) 4 5(12) 3 − 6 + 1(12) 2 Reduce each fraction 2(4) 1(3) − 5(2) + 1(6) Multiply 8 3 − 10 + 6 5 16 Add and subtract Our Solution Clearly the second method is a much cleaner and faster method to arrive at our solution. It is the method we will use when simplifying with variables as well. We will ﬁrst ﬁnd the LCD of the small fractions, and multiply each term by this LCD so we can clear the small fractions and simplify. Example 355. 1 x2 1 x 1 1 − − Identify LCD (use highest exponent) LCD = x2 Multiply each term by LCD 1(x2) 1(x2) 1(x2) x2 1(x2) x − − 1(x2) 1(x2) x2 x2 1 x 1 x − − − − Reduce fractions (subtract exponents) Multiply Factor (x + 1)(x x(x − 1) − 1) Divide out (x 1) factor − x + 1 x Our Solution The process is the same if the LCD is a binomial, we will need to distribute Multiply each term by LCD, (x + 4) 263 3(x + 4) 2(
x + 4) x + 4 − 5(x + 4) + 2(x + 4) x + 4 Reduce fractions 2(x + 4) 3 5(x + 4) + 2 − Distribute 2x 3 8 5x + 20 + 2 − − Combine like terms 2x 5 − − 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 − 4 a2b + ab ab 1 ab − Idenﬁty LCD (highest exponents) LCD = a2b3 Multiply each term by LCD ab3 + 1(a2b3) 3(a2b3) 2(a2b3) ab2 − 4(a2b3) a2b + ab(a2b3) ab 1(a2b3) ab − Reduce each fraction (subtract exponents) 3a + ab2 2ab − 4b2 + a3b4 ab2 Our Solution − World View Note: Sophie Germain is one of the most famous women in mathematics, many primes, which are important to ﬁnding an LCD, carry her name. Germain primes are prime numbers where one more than double the prime 3 + 1 = 7 prime. The number is also prime, for example 3 is prime and so is 2 largest known Germain prime (at the time of printing) is 183027 1 which has 79911 digits! 2265440 − · · Some problems may require us to FOIL as we simplify. To avoid sign errors, if there is a binomial in the numerator, we will ﬁrst distribute the negative through the numerator. Example 357 − Distribute the subtraction to numerator Identify LCD 264 LCD = (x + 3)(x − 3) Multiply each term by LCD (x − 3)(x + 3)(x x + 3 − 3) (x − 3)(x + 3)(x x + 3 − x − 3)(x + 3)(x 3 + ( − + (x + 3)(x + 3)(x 3) − x 3) − 3) − x 3 − (x − (x 3)(x − 3)(x 3) + ( x 3)(x + 3) − 3) + (x + 3)(x + 3) − − − x2 x2 − − x
2 6x + 9 6x − 6x + 9 + x2 + 6x − 9 9 − − 12x − 2x2 + 18 12x 2(x2 + 9) − − x2 6x 9 − 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 ﬁrst convert these negative exponents into fractions. Remember, the exponent is only on the factor it is attached to, not the whole term. Example 358. 1 2 + 2m− m− m + 4m− 2 1 m2 + 2 m m + 4 m2 1(m2) m2 + 2(m2) m(m2) + 4(m2) m2 m Make each negative exponent into a fraction Multiply each term by LCD, m2 Reduce the fractions 1 + 2m m3 + 4 Our Solution Once we convert each negative exponent into a fraction, the problem solves exactly like the other complex fraction problems. 265 7.5 Practice - Complex Fractions Solve. 1) 3) 5) 7) 9) 11) 13) 15) 17) 19) 21) 1 + 1 x 1 x2 − a2 − a2 + 1 1 a 2 5 − − 3 2a − 6 − − 2a 4 x + 2 10 x2 a2 b2 − 4a2b a + b 16ab2 1 − 1 + 11 10 3 x2 x − x + 18 x2 1 x 2x − 3x − 32 3x − − y2 − 1 + 1 y 25 b2 1 − 4 + 12 2x − 5 + 15 2x − − 10 b − 2a 3x − x 9x2 4 − 6 1 x2 x − x + 3 4 x2 1 1 − − 15 x2 − 4 x2 − 1 2 x − 5 x + 4 12 3x + 10 8 3x + 10 1 x − − − 18 ) 4) 6) 8) 10) 12) 14) 16) 18) 20) 22) 266 23) 25) 27) 29) x − x + 3 4 + 9 2x + 3 5 2x + b2 − 2 b2 + 7 5 3 a2 ab − ab + 3 a2 24) 26) 28) 30 y2 − 1 y2 − 1 2 x2 xy − xy + 2
3 x2 x + 1)2 + 1 (x 1)2 − Simplify each of the following fractional expressions. 31) x− x− 2 y− − 1 + y− 2 1 33) x− 3y x− 2 − − 3 xy− y− 2 2 x− 35) 1 + 9 6x− 9 − x2 − 32) 2 2y + xy− 2 2 x− x− y− − 2 3 4x− 4 1 + x− x− 2 − x− x− 3 + y− 1y− − 1 + y− 2 4 − 34) 36) 2 x− 267 7.6 Rational Expressions - Proportions Objective: Solve proportions using the cross product and use proportions to solve application problems When two fractions are equal, they are called a proportion. This deﬁnition can be generalized to two equal rational expressions. Proportions have an important property called the cross-product. Cross Product: If a b = c d then ad = bc The cross product tells us we can multiply diagonally to get an equation with no fractions that we can solve. Example 359. 20 6 = x 9 Calculate cross product 268 (20)(9) = 6x Multiply 180 = 6x Divide both sides by 6 6 30 = x Our Solution 6 World View Note: The ﬁrst clear deﬁnition of a proportion and the notation for a proportion came from the German Leibniz who wrote, “I write dy: x = dt: a; for dy is to x as dt is to a, is indeed the same as, dy divided by x is equal to dt divided by a. From this equation follow then all the rules of proportion.” If the proportion has more than one term in either numerator or denominator, we will have to distribute while calculating the cross product. Example 360. x + 3 4 = 2 5 Calculate cross product 5(x + 3) = (4)(2) Multiply and distribute Solve Subtract 15 from both sides 5x + 15 = 8 15 15 − − 5x = 7 Divide both sides by 5 − 5 5 7 5 Our Solution x = − This same idea can be seen when the variable appears in several parts of the proportion. Example 361. 4 x = 6 3x + 2 Calculate cross product 4(3x + 2) = 6x Dist

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