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Rational Expressions: Building Rational Expressions and the LCD

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary:

This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr.

A detailed study of arithmetic operations with rational expressions is presented in this chapter, beginning with the definition of a rational expression and then proceeding immediately to a discussion of the domain. The process of reducing a rational expression and illustrations of multiplying, dividing, adding, and subtracting rational expressions are also included. Since the operations of addition and subtraction can cause the most difficulty, they are given particular attention. We have tried to make the written explanation of the examples clearer by using a "freeze frame" approach, which walks the student through the operation step by step.

The five-step method of solving applied problems is included in this chapter to show the problem-solving approach to number problems, work problems, and geometry problems. The chapter also illustrates simplification of complex rational expressions, using the combine-divide method and the LCD-multiply-divide method.

Objectives of this module: understand and be able to use the process of building rational expressions and know why it is often necessary to build them, be able to find the LCD of one or more expressions.

Overview

  • The Process
  • The Reason For Building Rational Expressions
  • The Least Common Denominator (LCD)

The Process

Recall, from Section (Reference), the equality property of fractions.

Equality Property Of Fractions

If a b = c d , a b = c d , then ad=bc. ad=bc.

Using the fact that 1= b b ,b0 1= b b ,b0 , and that 1 is the multiplicative identity, it follows that if P Q P Q is a rational expression, then

P Q · b b = Pb Qb , b0 P Q · b b = Pb Qb , b0

This equation asserts that a rational expression can be transformed into an equivalent rational expression by multiplying both the numerator and denominator by the same nonzero number.

Process of Building Rational Expressions

This process is known as the process of building rational expressions and it is exactly the opposite of reducing rational expressions. The process is shown in these examples:

Example 1

3 4 3 4 can be built to 12 16 12 16 since

3 4 ·1= 3 4 · 4 4 = 3·4 4·4 = 12 16 3 4 ·1= 3 4 · 4 4 = 3·4 4·4 = 12 16

Example 2

4 5 4 5 can be built to 8 10 8 10 since

4 5 ·1= 4 5 · 2 2 = 4·2 5·2 = 8 10 4 5 ·1= 4 5 · 2 2 = 4·2 5·2 = 8 10

Example 3

3 7 3 7 can be built to 3xy 7xy 3xy 7xy since

3 7 ·1= 3 7 · xy xy = 3xy 7xy 3 7 ·1= 3 7 · xy xy = 3xy 7xy

Example 4

4a 3b 4a 3b can be built to 4 a 2 ( a+1 ) 3ab( a+1 ) 4 a 2 ( a+1 ) 3ab( a+1 ) since

4a 3b ·1= 4a 3b · a( a+1 ) a( a+1 ) = 4 a 2 ( a+1 ) 3ab( a+1 ) 4a 3b ·1= 4a 3b · a( a+1 ) a( a+1 ) = 4 a 2 ( a+1 ) 3ab( a+1 )

Suppose we're given a rational expression P Q P Q and wish to build it into a rational expression with denominator Q b 2 Q b 2 , that is,

P Q ? Q b 2 P Q ? Q b 2

Since we changed the denominator, we must certainly change the numerator in the same way. To determine how to change the numerator we need to know how the denominator was changed. Since one rational expression is built into another equivalent expression by multiplication by 1, the first denominator must have been multiplied by some quantity. Observation of

P Q ? Q b 2 P Q ? Q b 2

tells us that Q Q was multiplied by b 2 b 2 . Hence, we must multiply the numerator P P by b 2 b 2 . Thus,

P Q = P b 2 Q b 2 P Q = P b 2 Q b 2

Quite often a simple comparison of the original denominator with the new denominator will tell us the factor being used. However, there will be times when the factor is unclear by simple observation. We need a method for finding the factor.

Observe the following examples; then try to speculate on the method.

Example 5

3 4 = ? 20 . 3 4 = ? 20 .

The original denominator 4 was multiplied by 5 to yield 20. What arithmetic process will yield 5 using 4 and 20?

Example 6

9 10 = ? 10y . 9 10 = ? 10y .

The original denominator 10 was multiplied by y y to yield 10y 10y .

Example 7

6xy 2 a 3 b = ? 16 a 5 b 3 . 6xy 2 a 3 b = ? 16 a 5 b 3 .

The original denominator 2 a 3 b 2 a 3 b was multiplied by 8 a 2 b 2 8 a 2 b 2 to yield 16 a 5 b 3 . 16 a 5 b 3 .

Example 8

5ax ( a+1 ) 2 = ? 4 ( a+1 ) 2 ( a2 ) . 5ax ( a+1 ) 2 = ? 4 ( a+1 ) 2 ( a2 ) .

The original denominator ( a+1 ) 2 ( a+1 ) 2 was multiplied by 4( a2 ) 4( a2 ) to yield 4 ( a+1 ) 2 ( a2 ) 4 ( a+1 ) 2 ( a2 ) .

To determine the quantity that the original denominator was multiplied by to yield the new denominator, we ask, "What did I multiply the original denominator by to get the new denominator?" We find this factor by dividing the original denominator into the new denominator.

It is precisely this quantity that we multiply the numerator by to build the rational expression.

Sample Set A

Determine N in each of the following problems.

Example 9

8 3 = N 15 . The original denominator is 3 and the new  denominato is 15. Divide the original  denominator into the new denominator and  multiply the numerator 8 by this result. 15÷3=5 Then,85=40.So, 8 3 = 40 15 andN=40. Check by reducing 40 15 . 8 3 = N 15 . The original denominator is 3 and the new  denominato is 15. Divide the original  denominator into the new denominator and  multiply the numerator 8 by this result. 15÷3=5 Then,85=40.So, 8 3 = 40 15 andN=40. Check by reducing 40 15 .

Example 10

2x 5 b 2 y = N 20 b 5 y 4 . The original denominator is 5 b 2 y and the new  denominator is 20 b 5 y 4 . Divide the original  denominator into the new denominator and  multiply the numerator 2x by this result. 20 b 5 y 4 5 b 2 y =4 b 3 y 3 So,2x·4 b 3 y 3 =8 b 3 x y 3 .Thus, 2x 5 b 2 y = 8 b 3 x y 3 20 b 5 y 4 andN=8 b 3 x y 3 . 2x 5 b 2 y = N 20 b 5 y 4 . The original denominator is 5 b 2 y and the new  denominator is 20 b 5 y 4 . Divide the original  denominator into the new denominator and  multiply the numerator 2x by this result. 20 b 5 y 4 5 b 2 y =4 b 3 y 3 So,2x·4 b 3 y 3 =8 b 3 x y 3 .Thus, 2x 5 b 2 y = 8 b 3 x y 3 20 b 5 y 4 andN=8 b 3 x y 3 .

Example 11

6a a+2 = N ( a+2 )( a7 ) . The new denominator divided by the original denominator is  ( a+2 )( a7 ) a+2 =a7 Multiply 6abya7. 6a( a7 )=6 a 2 +42a 6a a+2 = 6 a 2 +42a ( a+2 )( a7 )  and N=6 a 2 +42a. 6a a+2 = N ( a+2 )( a7 ) . The new denominator divided by the original denominator is  ( a+2 )( a7 ) a+2 =a7 Multiply 6abya7. 6a( a7 )=6 a 2 +42a 6a a+2 = 6 a 2 +42a ( a+2 )( a7 )  and N=6 a 2 +42a.

Example 12

3( a1 ) a4 = N a 2 16 . The new denominator divided by the original denominator is a 2 16 a4    = ( a+4 ) ( a4 ) a4 =a+4 Multiply 3( a1 )bya+4. 3( a1 )( a+4 ) = 3( a 2 +3a4 ) = 3 a 2 9a+12 3( a1 ) a4 = 3 a 2 9a+12 a 2 16 andN=3 a 2 9a+12 3( a1 ) a4 = N a 2 16 . The new denominator divided by the original denominator is a 2 16 a4    = ( a+4 ) ( a4 ) a4 =a+4 Multiply 3( a1 )bya+4. 3( a1 )( a+4 ) = 3( a 2 +3a4 ) = 3 a 2 9a+12 3( a1 ) a4 = 3 a 2 9a+12 a 2 16 andN=3 a 2 9a+12

Example 13

7x = N x 2 y 3 . Write 7x as  7x 1 . 7x 1 = N x 2 y 3 Now we can see clearly that the original denominator  1 was multiplied by  x 2 y 3 . We need to multiply the  numerator 7x by  x 2 y 3 . 7x = 7x· x 2 y 3 x 2 y 3 7x = 7 x 3 y 3 x 2 y 3  and N=7 x 3 y 3 . 7x = N x 2 y 3 . Write 7x as  7x 1 . 7x 1 = N x 2 y 3 Now we can see clearly that the original denominator  1 was multiplied by  x 2 y 3 . We need to multiply the  numerator 7x by  x 2 y 3 . 7x = 7x· x 2 y 3 x 2 y 3 7x = 7 x 3 y 3 x 2 y 3  and N=7 x 3 y 3 .

Example 14

5x x+3 = 5 x 2 20x N . The same process works in this case. Divide the original  numerator 5x into the new numerator 5 x 2 20x. 5 x 2 20x 5x = 5x ( x4 ) 5x =x4 Multiply the denominator by x4. ( x+3 )( x4 ) 5x x+3 = 5 x 2 20 ( x+3 )( x4 ) andN=5 x 2 20. 5x x+3 = 5 x 2 20x N . The same process works in this case. Divide the original  numerator 5x into the new numerator 5 x 2 20x. 5 x 2 20x 5x = 5x ( x4 ) 5x =x4 Multiply the denominator by x4. ( x+3 )( x4 ) 5x x+3 = 5 x 2 20 ( x+3 )( x4 ) andN=5 x 2 20.

Example 15

4x 3x = N x3 . The two denominators have nearly the same terms; each has  the opposite sign. Factor 1 from the original denominator. 3x=1( 3+x ) =( x3 ) 4x 3x = 4x ( x3 ) = 4x x3  and N=4x. 4x 3x = N x3 . The two denominators have nearly the same terms; each has  the opposite sign. Factor 1 from the original denominator. 3x=1( 3+x ) =( x3 ) 4x 3x = 4x ( x3 ) = 4x x3  and N=4x.

It is important to note that we factored 1 1 from the original denominator. We did not multiply it by 1 1 . Had we multiplied only the denominator by 1 1 we would have had to multiply the numerator by 1 1 also.

Practice Set A

Determine N.

Exercise 1

3 8 = N 48 3 8 = N 48

Solution

N=18 N=18

Exercise 2

9a 5b = N 35 b 2 x 3 9a 5b = N 35 b 2 x 3

Solution

N=63ab x 3 N=63ab x 3

Exercise 3

2y y1 = N y 2 1 2y y1 = N y 2 1

Solution

N=2 y 2 2y N=2 y 2 2y

Exercise 4

a+7 a5 = N a 2 3a10 a+7 a5 = N a 2 3a10

Solution

N= a 2 +9a+14 N= a 2 +9a+14

Exercise 5

4a= N 6 a 3 ( a1 ) 4a= N 6 a 3 ( a1 )

Solution

N=24 a 4 ( a1 ) N=24 a 4 ( a1 )

Exercise 6

2x= N 8 x 3 y 3 z 5 2x= N 8 x 3 y 3 z 5

Solution

N=16 x 4 y 3 z 5 N=16 x 4 y 3 z 5

Exercise 7

6ab b+3 = N b 2 +6b+9 6ab b+3 = N b 2 +6b+9

Solution

N=6a b 2 +18ab N=6a b 2 +18ab

Exercise 8

3m m+5 = 3 m 2 18m N 3m m+5 = 3 m 2 18m N

Solution

N= m 2 m30 N= m 2 m30

Exercise 9

2 r 2 r3 = 2 r 3 +8 r 2 N 2 r 2 r3 = 2 r 3 +8 r 2 N

Solution

N= r 2 7r+12 N= r 2 7r+12

Exercise 10

8a b 2 a4 = N 4a 8a b 2 a4 = N 4a

Solution

N=8a b 2 N=8a b 2

The Reason For Building Rational Expressions

Building Rational Expressions

Normally, when we write a rational expression, we write it in reduced form. The reason for building rational expressions is to make addition and subtraction of rational expressions convenient (simpler).

To add or subtract two or more rational expressions they must have the same denominator.

Building rational expressions allows us to transform fractions into fractions with the same denominators (which we can then add or subtract). The most convenient new denominator is the least common denominator (LCD) of the given fractions.

The Least Common Denominator (LCD)

In arithmetic, the least common denominator is the smallest (least) quantity that each of the given denominators will divide into without a remainder. For algebraic expressions, the LCD is the polynomial of least degree divisible by each denominator. Some examples are shown below.

Example 16

3 4 , 1 6 , 5 12 . 3 4 , 1 6 , 5 12 .

The LCD is 12 since 12 is the smallest number that 4, 6, and 12 will divide into without a remainder.

Example 17

1 3 , 5 6 , 5 8 , 7 12 . 1 3 , 5 6 , 5 8 , 7 12 .

The LCD is 24 since 24 is the smallest number that 3, 6, 8, and 12 will divide into without a remainder.

Example 18

2 x , 3 x 2 . 2 x , 3 x 2 .

The LCD is x 2 x 2 since x 2 x 2 is the smallest quantity that x x and x 2 x 2 will divide into without a remainder.

Example 19

5a 6 a 2 b , 3a 8a b 3 . 5a 6 a 2 b , 3a 8a b 3 .

The LCD is 24 a 2 b 3 24 a 2 b 3 since 24 a 2 b 3 24 a 2 b 3 is the smallest quantity that 6 a 2 b 6 a 2 b and 8a b 3 8a b 3 will divide into without a remainder.

Example 20

2y y6 , 4 y 2 ( y6 ) 3 , y y1 . 2y y6 , 4 y 2 ( y6 ) 3 , y y1 .

The LCD is ( y6 ) 3 ( y1 ) ( y6 ) 3 ( y1 ) since ( y6 ) 3 ·( y1 ) ( y6 ) 3 ·( y1 ) is the smallest quantity that y6, ( y6 ) 3 y6, ( y6 ) 3 and y1 y1 will divide into without a remainder.

We’ll now propose and demonstrate a method for obtaining the LCD.

Method for Obtaining the LCD

  1. Factor each denominator. Use exponents for repeated factors. It is usually not necessary to factor numerical quantities.
  2. Write down each different factor that appears. If a factor appears more than once, use only the factor with the highest exponent.
  3. The LCD is the product of the factors written in step 2.

Sample Set B

Find the LCD.

Example 21

  • 1 x , 3 x 3 , 2 4y 1 x , 3 x 3 , 2 4y
  1. The denominators are already factored.
  2. Note that x x appears as x x and x 3 x 3 . Use only the x x with the higher exponent, x 3 x 3 . The term 4y 4y appears, so we must also use 4y 4y .
  3. The LCD is 4 x 3 y 4 x 3 y .

Example 22

  • 5 ( x1 ) 2 , 2x ( x1 )( x4 ) , 5x x 2 3x+2 5 ( x1 ) 2 , 2x ( x1 )( x4 ) , 5x x 2 3x+2
  1. Only the third denominator needs to be factored.

    x 2 3x+2=( x2 )( x1 ) x 2 3x+2=( x2 )( x1 )

    Now the three denominators are ( x1 ) 2 ,( x1 )( x4 ) ( x1 ) 2 ,( x1 )( x4 ) , and ( x2 )( x1 ). ( x2 )( x1 ).
  2. Note that x1 x1 appears as ( x1 ) 2 ,x1 ( x1 ) 2 ,x1 , and x1. x1. Use only the x1 x1 with the highest exponent, ( x1 ) 2 ( x1 ) 2 . Also appearing are x4 x4 and x2. x2.
  3. The LCD is ( x1 ) 2 ( x4 )( x2 ). ( x1 ) 2 ( x4 )( x2 ).

Example 23

  • 1 6 a 4 , 3 4 a 3 b , 1 3 a 3 ( b+5 ) 1 6 a 4 , 3 4 a 3 b , 1 3 a 3 ( b+5 )
  1. The denominators are already factored.
  2. We can see that the LCD of the numbers 6, 4, and 3 is 12. We also need a 4 a 4 , b b , and b+5 b+5 .
  3. The LCD is 12 a 4 b( b+5 ). 12 a 4 b( b+5 ).

Example 24

  • 9 x , 4 8y 9 x , 4 8y
  1. The denominators are already factored.
  2. x,8y. x,8y.
  3. The LCD is 8xy 8xy .

Practice Set B

Find the LCD.

Exercise 11

3 x 2 , 4 x 5 , 6 xy 3 x 2 , 4 x 5 , 6 xy

Solution

x 5 y x 5 y

Exercise 12

x+1 x4 , x7 ( x4 ) 2 , 6 x+1 x+1 x4 , x7 ( x4 ) 2 , 6 x+1

Solution

( x4 ) 2 ( x+1 ) ( x4 ) 2 ( x+1 )

Exercise 13

2 m6 , 5m ( m+1 ) 2 ( m2 ) , 12 m 2 ( m2 ) 3 ( m6 ) 2 m6 , 5m ( m+1 ) 2 ( m2 ) , 12 m 2 ( m2 ) 3 ( m6 )

Solution

( m6 ) ( m+1 ) 2 ( m2 ) 3 ( m6 ) ( m+1 ) 2 ( m2 ) 3

Exercise 14

1 x 2 1 , 2 x 2 2x3 , 3x x 2 6x+9 1 x 2 1 , 2 x 2 2x3 , 3x x 2 6x+9

Solution

( x+1 )( x1 ) ( x3 ) 2 ( x+1 )( x1 ) ( x3 ) 2

Exercise 15

3 4 y 2 8y , 8 y 2 4y+4 , 10y1 3 y 3 6 y 2 3 4 y 2 8y , 8 y 2 4y+4 , 10y1 3 y 3 6 y 2

Solution

12 y 2 ( y2 ) 2 12 y 2 ( y2 ) 2

Sample Set C

Change the given rational expressions into rational expressions having the same denominator.

Example 25

3 x 2 , 4 x . The LCD, by inspection, is  x 2 . Rewrite each expression  with  x 2  as the new denominator.  x 2 , x 2 Determine the numerators. In  3 x 2 , the denominator was not  changed so we need not change the numerator.  3 x 2 , x 2 In the second fraction, the original denominator was x.  We can see that x must be multiplied by x to build it to  x 2 .  So we must also multiply the numerator 4 by x. Thus, 4·x=4x.  3 x 2 , 4x x 2 3 x 2 , 4 x . The LCD, by inspection, is  x 2 . Rewrite each expression  with  x 2  as the new denominator.  x 2 , x 2 Determine the numerators. In  3 x 2 , the denominator was not  changed so we need not change the numerator.  3 x 2 , x 2 In the second fraction, the original denominator was x.  We can see that x must be multiplied by x to build it to  x 2 .  So we must also multiply the numerator 4 by x. Thus, 4·x=4x.  3 x 2 , 4x x 2

Example 26

4b b1 , 2b b+3 . By inspection, the LCD is ( b1 )( b+3 ). Rewrite each fraction with new denominator ( b1 )( b+3 ). ( b1 )( b+3 ) , ( b1 )( b+3 ) The denominator of the first rational expression has been multiplied  by b+3,so the numerator 4b must be multiplied by b+3. 4b( b+3 )=4 b 2 +12b 4 b 2 +12b ( b1 )( b+3 ) , ( b1 )( b+3 ) The denominator of the second rational expression has been multiplied  by b1, so the numerator 2b must be multiplied by b1. 2b( b1 )=2 b 2 +2b 4 b 2 +12b ( b1 )( b+3 ) , 2 b 2 +2b ( b1 )( b+3 ) 4b b1 , 2b b+3 . By inspection, the LCD is ( b1 )( b+3 ). Rewrite each fraction with new denominator ( b1 )( b+3 ). ( b1 )( b+3 ) , ( b1 )( b+3 ) The denominator of the first rational expression has been multiplied  by b+3,so the numerator 4b must be multiplied by b+3. 4b( b+3 )=4 b 2 +12b 4 b 2 +12b ( b1 )( b+3 ) , ( b1 )( b+3 ) The denominator of the second rational expression has been multiplied  by b1, so the numerator 2b must be multiplied by b1. 2b( b1 )=2 b 2 +2b 4 b 2 +12b ( b1 )( b+3 ) , 2 b 2 +2b ( b1 )( b+3 )

Example 27

6x x 2 8x+15 , 2 x 2 x 2 7x+12 . We first find the LCD. Factor. 6x ( x3 )( x5 ) , 2 x 2 ( x3 )( x4 ) The LCD is ( x3 )( x5 )( x4 ).Rewrite each of these  fractions with new denominator ( x3 )( x5 )( x4 ). ( x3 )( x5 )( x4 ) , ( x3 )( x5 )( x4 ) By comparing the denominator of the first fraction with the LCD  we see that we must multiply the numerator 6x by x4. 6x( x4 )=6 x 2 24x 6 x 2 24x ( x3 )( x5 )( x4 ) , ( x3 )( x5 )( x4 ) By comparing the denominator of the second fraction with the LCD,  we see that we must multiply the numerator 2 x 2  by x5. 2 x 2 ( x5 )=2 x 3 +10 x 2 6 x 2 24x ( x3 )( x5 )( x4 ) , 2 x 3 +10 x 2 ( x3 )( x5 )( x4 ) 6x x 2 8x+15 , 2 x 2 x 2 7x+12 . We first find the LCD. Factor. 6x ( x3 )( x5 ) , 2 x 2 ( x3 )( x4 ) The LCD is ( x3 )( x5 )( x4 ).Rewrite each of these  fractions with new denominator ( x3 )( x5 )( x4 ). ( x3 )( x5 )( x4 ) , ( x3 )( x5 )( x4 ) By comparing the denominator of the first fraction with the LCD  we see that we must multiply the numerator 6x by x4. 6x( x4 )=6 x 2 24x 6 x 2 24x ( x3 )( x5 )( x4 ) , ( x3 )( x5 )( x4 ) By comparing the denominator of the second fraction with the LCD,  we see that we must multiply the numerator 2 x 2  by x5. 2 x 2 ( x5 )=2 x 3 +10 x 2 6 x 2 24x ( x3 )( x5 )( x4 ) , 2 x 3 +10 x 2 ( x3 )( x5 )( x4 )

These examples have been done step-by-step and include explanations. This makes the process seem fairly long. In practice, however, the process is much quicker.

Example 28

6ab a 2 5a+4 , a+b a 2 8a+16 6ab ( a1 )( a4 ) , a+b ( a4 ) 2 LCD=( a1 ) ( a4 ) 2 . 6ab( a4 ) ( a1 ) ( a4 ) 2 , ( a+b )( a1 ) ( a1 ) ( a4 ) 2 6ab a 2 5a+4 , a+b a 2 8a+16 6ab ( a1 )( a4 ) , a+b ( a4 ) 2 LCD=( a1 ) ( a4 ) 2 . 6ab( a4 ) ( a1 ) ( a4 ) 2 , ( a+b )( a1 ) ( a1 ) ( a4 ) 2

Example 29

x+1 x 3 +3 x 2 , 2x x 3 4x , x4 x 2 4x+4 x+1 x 2 ( x+3 ) , 2x x( x+2 )( x2 ) , x4 ( x2 ) 2 LCD= x 2 ( x+3 )( x+2 ) ( x2 ) 2 . ( x+1 )( x+2 ) ( x2 ) 2 x 2 ( x+3 )( x+2 ) ( x2 ) 2 , 2 x 2 ( x+3 )( x2 ) x 2 ( x+3 )( x+2 ) ( x2 ) 2 , x 2 ( x+3 )( x+2 )( x4 ) x 2 ( x+3 )( x+2 ) ( x2 ) 2 x+1 x 3 +3 x 2 , 2x x 3 4x , x4 x 2 4x+4 x+1 x 2 ( x+3 ) , 2x x( x+2 )( x2 ) , x4 ( x2 ) 2 LCD= x 2 ( x+3 )( x+2 ) ( x2 ) 2 . ( x+1 )( x+2 ) ( x2 ) 2 x 2 ( x+3 )( x+2 ) ( x2 ) 2 , 2 x 2 ( x+3 )( x2 ) x 2 ( x+3 )( x+2 ) ( x2 ) 2 , x 2 ( x+3 )( x+2 )( x4 ) x 2 ( x+3 )( x+2 ) ( x2 ) 2

Practice Set C

Change the given rational expressions into rational expressions with the same denominators.

Exercise 16

4 x 3 , 7 x 5 4 x 3 , 7 x 5

Solution

4 x 2 x 5 , 7 x 5 4 x 2 x 5 , 7 x 5

Exercise 17

2x x+6 , x x1 2x x+6 , x x1

Solution

2x( x1 ) ( x+6 )( x1 ) , x( x+6 ) ( x+6 )( x1 ) 2x( x1 ) ( x+6 )( x1 ) , x( x+6 ) ( x+6 )( x1 )

Exercise 18

3 b 2 b , 4b b 2 1 3 b 2 b , 4b b 2 1

Solution

3( b+1 ) b( b1 )( b+1 ) , 4 b 2 b( b1 )( b+1 ) 3( b+1 ) b( b1 )( b+1 ) , 4 b 2 b( b1 )( b+1 )

Exercise 19

8 x 2 x6 , 1 x 2 +x2 8 x 2 x6 , 1 x 2 +x2

Solution

8( x1 ) ( x3 )( x+2 )( x1 ) , 1( x3 ) ( x3 )( x+2 )( x1 ) 8( x1 ) ( x3 )( x+2 )( x1 ) , 1( x3 ) ( x3 )( x+2 )( x1 )

Exercise 20

10x x 2 +8x+16 , 5x x 2 16 10x x 2 +8x+16 , 5x x 2 16

Solution

10x( x4 ) ( x+4 ) 2 ( x4 ) , 5x( x+4 ) ( x+4 ) 2 ( x4 ) 10x( x4 ) ( x+4 ) 2 ( x4 ) , 5x( x+4 ) ( x+4 ) 2 ( x4 )

Exercise 21

2a b 2 a 3 6 a 2 , 6b a 4 2 a 3 , 2a a 2 4a+4 2a b 2 a 3 6 a 2 , 6b a 4 2 a 3 , 2a a 2 4a+4

Solution

2 a 2 b 2 ( a2 ) 2 a 3 ( a6 ) ( a2 ) 2 , 6b( a6 )( a2 ) a 3 ( a6 ) ( a2 ) 2 , 2 a 4 ( a6 ) a 3 ( a6 ) ( a2 ) 2 2 a 2 b 2 ( a2 ) 2 a 3 ( a6 ) ( a2 ) 2 , 6b( a6 )( a2 ) a 3 ( a6 ) ( a2 ) 2 , 2 a 4 ( a6 ) a 3 ( a6 ) ( a2 ) 2

Exercises

For the following problems, replace N N with the proper quantity.

Exercise 22

3x=Nx33x=Nx3

Solution

3 x 2 3 x 2

Exercise 23

4a=Na24a=Na2

Exercise 24

2x=Nxy2x=Nxy

Solution

2y 2y

Exercise 25

7m=Nms7m=Nms

Exercise 26

6a5=N10b6a5=N10b

Solution

12ab 12ab

Exercise 27

a3z=N12za3z=N12z

Exercise 28

x24y2=N20y4x24y2=N20y4

Solution

5 x 2 y 2 5 x 2 y 2

Exercise 29

b36a=N18a5b36a=N18a5

Exercise 30

4a5x2y=N15x3y34a5x2y=N15x3y3

Solution

12ax y 2 12ax y 2

Exercise 31

10z7a3b=N21a4b510z7a3b=N21a4b5

Exercise 32

8x2y5a3=N25a3x28x2y5a3=N25a3x2

Solution

40 x 4 y 40 x 4 y

Exercise 33

2a2=Na2(a1)2a2=Na2(a1)

Exercise 34

5x3=Nx3(x2)5x3=Nx3(x2)

Solution

5( x2 ) 5( x2 )

Exercise 35

2ab2=Nb3b2ab2=Nb3b

Exercise 36

4xa=Na44a24xa=Na44a2

Solution

4ax( a+2 )( a2 ) 4ax( a+2 )( a2 )

Exercise 37

6b35a=N10a230a6b35a=N10a230a

Exercise 38

4x3b=N3b515b4x3b=N3b515b

Solution

4x( b 4 5 ) 4x( b 4 5 )

Exercise 39

2mm1=N(m1)(m+2)2mm1=N(m1)(m+2)

Exercise 40

3ss+12=N(s+12)(s7)3ss+12=N(s+12)(s7)

Solution

3s( s7 ) 3s( s7 )

Exercise 41

a+1a3=N(a3)(a4)a+1a3=N(a3)(a4)

Exercise 42

a+2a2=N(a2)(a4)a+2a2=N(a2)(a4)

Solution

( a+2 )( a4 ) ( a+2 )( a4 )

Exercise 43

b+7b6=N(b6)(b+6)b+7b6=N(b6)(b+6)

Exercise 44

5m2m+1=N(2m+1)(m2)5m2m+1=N(2m+1)(m2)

Solution

5m( m2 ) 5m( m2 )

Exercise 45

4a+6=Na2+5a64a+6=Na2+5a6

Exercise 46

9b2=Nb26b+89b2=Nb26b+8

Solution

9( b4 ) 9( b4 )

Exercise 47

3bb3=Nb211b+243bb3=Nb211b+24

Exercise 48

2xx7=Nx24x212xx7=Nx24x21

Solution

2x( x+3 ) 2x( x+3 )

Exercise 49

6mm+6=Nm2+10m+246mm+6=Nm2+10m+24

Exercise 50

4yy+1=Ny2+9y+84yy+1=Ny2+9y+8

Solution

4y( y+8 ) 4y( y+8 )

Exercise 51

x+2x2=Nx24x+2x2=Nx24

Exercise 52

y3y+3=Ny29y3y+3=Ny29

Solution

( y3 ) 2 ( y3 ) 2

Exercise 53

a+5 a5 = N a 2 25 a+5 a5 = N a 2 25

Exercise 54

z4z+4=Nz216z4z+4=Nz216

Solution

( z4 ) 2 ( z4 ) 2

Exercise 55

42a+1=N2a25a342a+1=N2a25a3

Exercise 56

13b1=N3b2+11b413b1=N3b2+11b4

Solution

b+4 b+4

Exercise 57

a+22a1=N2a2+9a5a+22a1=N2a2+9a5

Exercise 58

34x+3=N4x213x1234x+3=N4x213x12

Solution

3( x4 ) 3( x4 )

Exercise 59

b+23b1=N6b2+7b3b+23b1=N6b2+7b3

Exercise 60

x14x5=N12x211x5x14x5=N12x211x5

Solution

( x1 )( 3x+1 ) ( x1 )( 3x+1 )

Exercise 61

3x+2=3x21N3x+2=3x21N

Exercise 62

4y+6=4y+8N4y+6=4y+8N

Solution

( y+6 )( y+2 ) ( y+6 )( y+2 )

Exercise 63

6a1=6a18N6a1=6a18N

Exercise 64

8aa+3=8a240aN8aa+3=8a240aN

Solution

( a+3 )( a+5 ) ( a+3 )( a+5 )

Exercise 65

y+1y8=y22y3Ny+1y8=y22y3N

Exercise 66

x4x+9=x2+x20Nx4x+9=x2+x20N

Solution

( x+9 )( x+5 ) ( x+9 )( x+5 )

Exercise 67

3x2x=Nx23x2x=Nx2

Exercise 68

7a5a=Na57a5a=Na5

Solution

7a 7a

Exercise 69

m+13m=Nm3m+13m=Nm3

Exercise 70

k+610k=Nk10k+610k=Nk10

Solution

k6 k6

For the following problems, convert the given rational expressions to rational expressions having the same denominators.

Exercise 71

2a,3a42a,3a4

Exercise 72

5b2,4b35b2,4b3

Solution

5b b 3 , 4 b 3 5b b 3 , 4 b 3

Exercise 73

8z,34z38z,34z3

Exercise 74

9x2,14x9x2,14x

Solution

36 4 x 2 , x 4 x 2 36 4 x 2 , x 4 x 2

Exercise 75

2a+3,4a+12a+3,4a+1

Exercise 76

2x+5,4x52x+5,4x5

Solution

2( x5 ) ( x+5 )( x5 ) , 4( x+5 ) ( x+5 )( x5 ) 2( x5 ) ( x+5 )( x5 ) , 4( x+5 ) ( x+5 )( x5 )

Exercise 77

1x7,4x11x7,4x1

Exercise 78

10y+2,1y+810y+2,1y+8

Solution

10( y+8 ) ( y+2 )( y+8 ) , y+2 ( y+2 )( y+8 ) 10( y+8 ) ( y+2 )( y+8 ) , y+2 ( y+2 )( y+8 )

Exercise 79

4a2,aa+44a2,aa+4

Exercise 80

3b2,b2b+53b2,b2b+5

Solution

3( b+5 ) b 2 ( b+5 ) , b 4 b 2 ( b+5 ) 3( b+5 ) b 2 ( b+5 ) , b 4 b 2 ( b+5 )

Exercise 81

6b1,5b4b6b1,5b4b

Exercise 82

10aa6,2a26a10aa6,2a26a

Solution

10 a 2 a( a6 ) , 2 a( a6 ) 10 a 2 a( a6 ) , 2 a( a6 )

Exercise 83

4 x 2 +2x , 1 x 2 4 4 x 2 +2x , 1 x 2 4

Exercise 84

x+1 x 2 x6 , x+4 x 2 +x2 x+1 x 2 x6 , x+4 x 2 +x2

Solution

( x+1 )( x1 ) ( x1 )( x+2 )( x3 ) , ( x+4 )( x3 ) ( x1 )( x+2 )( x3 ) ( x+1 )( x1 ) ( x1 )( x+2 )( x3 ) , ( x+4 )( x3 ) ( x1 )( x+2 )( x3 )

Exercise 85

x5 x 2 9x+20 , 4 x 2 3x10 x5 x 2 9x+20 , 4 x 2 3x10

Exercise 86

4 b 2 +5b6 , b+6 b 2 1 4 b 2 +5b6 , b+6 b 2 1

Solution

4( b+1 ) ( b+1 )( b1 )( b+6 ) , ( b+6 ) 2 ( b+1 )( b1 )( b+6 ) 4( b+1 ) ( b+1 )( b1 )( b+6 ) , ( b+6 ) 2 ( b+1 )( b1 )( b+6 )

Exercise 87

b+2 b 2 +6b+8 , b1 b 2 +8b+12 b+2 b 2 +6b+8 , b1 b 2 +8b+12

Exercise 88

x+7 x 2 2x3 , x+3 x 2 6x7 x+7 x 2 2x3 , x+3 x 2 6x7

Solution

( x+7 )( x7 ) ( x+1 )( x3 )( x7 ) , ( x+3 )( x3 ) ( x+1 )( x3 )( x7 ) ( x+7 )( x7 ) ( x+1 )( x3 )( x7 ) , ( x+3 )( x3 ) ( x+1 )( x3 )( x7 )

Exercise 89

2 a 2 +a , a+3 a 2 1 2 a 2 +a , a+3 a 2 1

Exercise 90

x2 x 2 +7x+6 , 2x x 2 +4x12 x2 x 2 +7x+6 , 2x x 2 +4x12

Solution

( x2 ) 2 ( x+1 )( x2 )( x+6 ) , 2x( x+1 ) ( x+1 )( x2 )( x+6 ) ( x2 ) 2 ( x+1 )( x2 )( x+6 ) , 2x( x+1 ) ( x+1 )( x2 )( x+6 )

Exercise 91

x2 2 x 2 +5x3 , x.1 5 x 2 +16x+3 x2 2 x 2 +5x3 , x.1 5 x 2 +16x+3

Exercise 92

2 x5 , 3 5x 2 x5 , 3 5x

Solution

2 x5 , 3 x5 2 x5 , 3 x5

Exercise 93

4 a6 , 5 6a 4 a6 , 5 6a

Exercise 94

6 2x , 5 x2 6 2x , 5 x2

Solution

6 x2 , 5 x2 6 x2 , 5 x2

Exercise 95

k 5k , 3k k5 k 5k , 3k k5

Exercise 96

2m m8 , 7 8m 2m m8 , 7 8m

Solution

2m m8 , 7 m8 2m m8 , 7 m8

Excercises For Review

Exercise 97

((Reference)) Factor m 2 x 3 +m x 2 +mx. m 2 x 3 +m x 2 +mx.

Exercise 98

((Reference)) Factor y 2 10y+21. y 2 10y+21.

Solution

( y7 )( y3 ) ( y7 )( y3 )

Exercise 99

((Reference)) Write the equation of the line that passes through the points ( 1,1 ) ( 1,1 ) and ( 4,2 ) ( 4,2 ) . Express the equation in slope-intercept form.

Exercise 100

((Reference)) Reduce y 2 y6 y3 . y 2 y6 y3 .

Solution

y+2 y+2

Exercise 101

((Reference)) Find the quotient: x 2 6x+9 x 2 x6 ÷ x 2 +2x15 x 2 +2x . x 2 6x+9 x 2 x6 ÷ x 2 +2x15 x 2 +2x .

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