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# Reducing Rational Expressions

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 reducing rational expressions.

## Overview

• The Logic Behind The Process
• The Process

## The Logic Behind The Process

When working with rational expressions, it is often best to write them in the simplest possible form. For example, the rational expression
x 2 -4 x 2 -6x+8 x 2 -4 x 2 -6x+8
can be reduced to the simpler expression x+2 x-4 x+2 x-4 for all x x except x=2,4 x=2,4 .

From our discussion of equality of fractions in Section (Reference), we know that a b = c d a b = c d when ad=bc ad=bc . This fact allows us to deduce that, if k0, ak bk = a b , k0, ak bk = a b , since akb=abk akb=abk (recall the commutative property of multiplication). But this fact means that if a factor (in this case, k k ) is common to both the numerator and denominator of a fraction, we may remove it without changing the value of the fraction.
ak bk = a k b k = a b ak bk = a k b k = a b

### Cancelling

The process of removing common factors is commonly called cancelling.

#### Example 1

16 40 16 40 can be reduced to 2 5 2 5 .   Process:

16 40 = 2·2·2·2 2·2·2·5 16 40 = 2·2·2·2 2·2·2·5

Remove the three factors of 1; 2 2 · 2 2 · 2 2 . 2 2 · 2 2 · 2 2 .

2 · 2 · 2 ·2 2 · 2 · 2 ·5 = 2 5 2 · 2 · 2 ·2 2 · 2 · 2 ·5 = 2 5

Notice that in 2 5 2 5 , there is no factor common to the numerator and denominator.

#### Example 2

111 148 111 148 can be reduced to 3 4 3 4 .   Process:

111 148 = 3·37 4·37 111 148 = 3·37 4·37

Remove the factor of 1; 37 37 37 37 .

3· 37 4· 37 3· 37 4· 37

3 4 3 4

Notice that in 3 4 3 4 , there is no factor common to the numerator and denominator.

#### Example 3

3 9 3 9 can be reduced to 1 3 1 3 .   Process:

3 9 = 3·1 3·3 3 9 = 3·1 3·3

Remove the factor of 1; 3 3 3 3 .

3 ·1 3 ·3 = 1 3 3 ·1 3 ·3 = 1 3

Notice that in 1 3 1 3 there is no factor common to the numerator and denominator.

#### Example 4

5 7 5 7 cannot be reduced since there are no factors common to the numerator and denominator.

Problems 1, 2, and 3 shown above could all be reduced. The process in each reduction included the following steps:

1. Both the numerator and denominator were factored.
2. Factors that were common to both the numerator and denominator were noted and removed by dividing them out.

We know that we can divide both sides of an equation by the same nonzero number, but why should we be able to divide both the numerator and denominator of a fraction by the same nonzero number? The reason is that any nonzero number divided by itself is 1, and that if a number is multiplied by 1, it is left unchanged.

Consider the fraction 6 24 6 24 . Multiply this fraction by 1. This is written 6 24 ·1 6 24 ·1 . But 1 can be rewritten as 1 6 1 6 1 6 1 6 .

6 24 1 6 1 6 = 6 1 6 24 1 6 = 1 4 6 24 1 6 1 6 = 6 1 6 24 1 6 = 1 4

The answer, 1 4 1 4 , is the reduced form. Notice that in 1 4 1 4 there is no factor common to both the numerator and denominator. This reasoning provides justification for the following rule.

#### Cancelling

Multiplying or dividing the numerator and denominator by the same nonzero number does not change the value of a fraction.

## The Process

We can now state a process for reducing a rational expression.

### Reducing a Rational Expression

1. Factor the numerator and denominator completely.
2. Divide the numerator and denominator by all factors they have in common, that is, remove all factors of 1.

### Reduced to Lowest Terms

1. A rational expression is said to be reduced to lowest terms when the numerator and denominator have no factors in common.

## Sample Set A

Reduce the following rational expressions.

### Example 5

15x 20x . Factor. 15x 20x = 5·3·x 5·2·2·x The factors that are common to both the numerator and denominator are 5 and  x. Divide each by  5x. 5 ·3· x 5 ·2·2· x = 3 4 ,x0 It is helpful to draw a line through the divided-out factors. 15x 20x . Factor. 15x 20x = 5·3·x 5·2·2·x The factors that are common to both the numerator and denominator are 5 and  x. Divide each by  5x. 5 ·3· x 5 ·2·2· x = 3 4 ,x0 It is helpful to draw a line through the divided-out factors.

### Example 6

x 2 4 x 2 6x+8 . Factor. ( x+2 )( x2 ) ( x2 )( x4 ) The factor that is common to both the numerator and denominator is  x2. Divide each by x2. ( x+2 ) ( x2 ) ( x2 ) ( x4 ) = x+2 x4 ,x2,4 x 2 4 x 2 6x+8 . Factor. ( x+2 )( x2 ) ( x2 )( x4 ) The factor that is common to both the numerator and denominator is  x2. Divide each by x2. ( x+2 ) ( x2 ) ( x2 ) ( x4 ) = x+2 x4 ,x2,4

The expression x2 x4 x2 x4 is the reduced form since there are no factors common to both the numerator and denominator. Although there is an x x in both, it is a common term, not a common factor, and therefore cannot be divided out.

CAUTION — This is a common error: x2 x4 = x 2 x 4 = 2 4 x2 x4 = x 2 x 4 = 2 4 is incorrect!

### Example 7

a+2b 6a+12b . Factor. a+2b 6( a+2b ) = a+2b 6 ( a+2b ) = 1 6 ,a-2b a+2b 6a+12b . Factor. a+2b 6( a+2b ) = a+2b 6 ( a+2b ) = 1 6 ,a-2b
Since a+2b a+2b is a common factor to both the numerator and denominator, we divide both by a+2b a+2b . Since ( a+2b ) ( a+2b ) =1 ( a+2b ) ( a+2b ) =1 , we get 1 in the numerator.

Sometimes we may reduce a rational expression by using the division rule of exponents.

### Example 8

8 x 2 y 5 4x y 2 . Factor and use the rule  a n a m = a nm . 8 x 2 y 5 4x y 2 = 222 22 x 21 y 52 = 2x y 3 ,x0,y0 8 x 2 y 5 4x y 2 . Factor and use the rule  a n a m = a nm . 8 x 2 y 5 4x y 2 = 222 22 x 21 y 52 = 2x y 3 ,x0,y0

### Example 9

10 x 3 a( x 2 36 ) 2 x 3 10 x 2 12x . Factor. 10 x 3 a( x 2 36 ) 2 x 3 10 x 2 12x = 52 x 3 a( x+6 )( x6 ) 2x( x 2 5x6 ) = 52 x 3 a( x+6 )( x6 ) 2x( x6 )( x+1 ) = 5 2 x 2 3 a( x+6 ) ( x6 ) 2 x ( x6 ) ( x+1 ) = 5 x 2 a( x+6 ) x1 ,x1,6 10 x 3 a( x 2 36 ) 2 x 3 10 x 2 12x . Factor. 10 x 3 a( x 2 36 ) 2 x 3 10 x 2 12x = 52 x 3 a( x+6 )( x6 ) 2x( x 2 5x6 ) = 52 x 3 a( x+6 )( x6 ) 2x( x6 )( x+1 ) = 5 2 x 2 3 a( x+6 ) ( x6 ) 2 x ( x6 ) ( x+1 ) = 5 x 2 a( x+6 ) x1 ,x1,6

### Example 10

x 2 x12 x 2 +2x+8 . Since it is most convenient to have the leading terms of a polynomial positive, factor out 1 from the denominator. x 2 x12 ( x 2 2x8 ) Rewrite this. x 2 x12 x 2 2x8 Factor. ( x4 ) ( x+3 ) ( x4 ) ( x+2 ) x+3 x+2 = ( x+3 ) x+2 = x3 x+2 ,x2,4 x 2 x12 x 2 +2x+8 . Since it is most convenient to have the leading terms of a polynomial positive, factor out 1 from the denominator. x 2 x12 ( x 2 2x8 ) Rewrite this. x 2 x12 x 2 2x8 Factor. ( x4 ) ( x+3 ) ( x4 ) ( x+2 ) x+3 x+2 = ( x+3 ) x+2 = x3 x+2 ,x2,4

### Example 11

ab ba . The numerator and denominator have the same terms but they occur with opposite signs. Factor 1 from the denominator. ab ( b+a ) = ab ( ab ) = ab ab =1,ab ab ba . The numerator and denominator have the same terms but they occur with opposite signs. Factor 1 from the denominator. ab ( b+a ) = ab ( ab ) = ab ab =1,ab

## Practice Set A

Reduce each of the following fractions to lowest terms.

30y 35y 30y 35y

6 7 6 7

### Exercise 2

x 2 9 x 2 +5x+6 x 2 9 x 2 +5x+6

x3 x+2 x3 x+2

### Exercise 3

x+2b 4x+8b x+2b 4x+8b

1 4 1 4

### Exercise 4

18 a 3 b 5 c 7 3a b 3 c 5 18 a 3 b 5 c 7 3a b 3 c 5

#### Solution

6 a 2 b 2 c 2 6 a 2 b 2 c 2

### Exercise 5

3 a 4 +75 a 2 2 a 3 16 a 2 +30a 3 a 4 +75 a 2 2 a 3 16 a 2 +30a

#### Solution

3a( a+5 ) 2( a3 ) 3a( a+5 ) 2( a3 )

### Exercise 6

x 2 5x+4 x 2 +12x32 x 2 5x+4 x 2 +12x32

x+1 x8 x+1 x8

2xy y2x 2xy y2x

−1

## Excercises

For the following problems, reduce each rational expression to lowest terms.

### Exercise 8

6 3x-12 6 3x-12

#### Solution

2 ( x4 ) 2 ( x4 )

8 4a-16 8 4a-16

### Exercise 10

9 3y-21 9 3y-21

#### Solution

3 ( y7 ) 3 ( y7 )

10 5x-5 10 5x-5

### Exercise 12

7 7x-14 7 7x-14

#### Solution

1 ( x2 ) 1 ( x2 )

6 6x-18 6 6x-18

### Exercise 14

2 y 2 8y 2 y 2 8y

1 4 y 1 4 y

### Exercise 15

4 x 3 2x 4 x 3 2x

### Exercise 16

16 a 2 b 3 2a b 2 16 a 2 b 3 2a b 2

8ab 8ab

### Exercise 17

20 a 4 b 4 4a b 2 20 a 4 b 4 4a b 2

### Exercise 18

( x+3 )( x-2 ) ( x+3 )( x+5 ) ( x+3 )( x-2 ) ( x+3 )( x+5 )

x2 x+5 x2 x+5

### Exercise 19

( y-1 )( y-7 ) ( y-1 )( y+6 ) ( y-1 )( y-7 ) ( y-1 )( y+6 )

### Exercise 20

( a+6 )( a-5 ) ( a-5 )( a+2 ) ( a+6 )( a-5 ) ( a-5 )( a+2 )

a+6 a+2 a+6 a+2

### Exercise 21

( m-3 )( m-1 ) ( m-1 )( m+4 ) ( m-3 )( m-1 ) ( m-1 )( m+4 )

### Exercise 22

( y-2 )( y-3 ) ( y-3 )( y-2 ) ( y-2 )( y-3 ) ( y-3 )( y-2 )

1

### Exercise 23

( x+7 )( x+8 ) ( x+8 )( x+7 ) ( x+7 )( x+8 ) ( x+8 )( x+7 )

### Exercise 24

-12 x 2 ( x+4 ) 4x -12 x 2 ( x+4 ) 4x

#### Solution

3x( x+4 ) 3x( x+4 )

### Exercise 25

-3 a 4 ( a-1 )( a+5 ) -2 a 3 ( a-1 )( a+9 ) -3 a 4 ( a-1 )( a+5 ) -2 a 3 ( a-1 )( a+9 )

### Exercise 26

6 x 2 y 5 ( x-1 )( x+4 ) -2xy( x+4 ) 6 x 2 y 5 ( x-1 )( x+4 ) -2xy( x+4 )

#### Solution

3x y 4 ( x1 ) 3x y 4 ( x1 )

### Exercise 27

22 a 4 b 6 c 7 ( a+2 )( a-7 ) 4c( a+2 )( a-5 ) 22 a 4 b 6 c 7 ( a+2 )( a-7 ) 4c( a+2 )( a-5 )

### Exercise 28

( x+10 ) 3 x+10 ( x+10 ) 3 x+10

#### Solution

( x+10 ) 2 ( x+10 ) 2

### Exercise 29

( y-6 ) 7 y-6 ( y-6 ) 7 y-6

### Exercise 30

( x-8 ) 2 ( x+6 ) 4 ( x-8 )( x+6 ) ( x-8 ) 2 ( x+6 ) 4 ( x-8 )( x+6 )

#### Solution

( x8 ) ( x+6 ) 3 ( x8 ) ( x+6 ) 3

### Exercise 31

( a+1 ) 5 ( a-1 ) 7 ( a+1 ) 3 ( a-1 ) 4 ( a+1 ) 5 ( a-1 ) 7 ( a+1 ) 3 ( a-1 ) 4

### Exercise 32

( y-2 ) 6 ( y-1 ) 4 ( y-2 ) 3 ( y-1 ) 2 ( y-2 ) 6 ( y-1 ) 4 ( y-2 ) 3 ( y-1 ) 2

#### Solution

( y2 ) 3 ( y1 ) 2 ( y2 ) 3 ( y1 ) 2

### Exercise 33

( x+10 ) 5 ( x-6 ) 3 ( x-6 ) ( x+10 ) 2 ( x+10 ) 5 ( x-6 ) 3 ( x-6 ) ( x+10 ) 2

### Exercise 34

( a+6 ) 2 ( a-7 ) 6 ( a+6 ) 5 ( a-7 ) 2 ( a+6 ) 2 ( a-7 ) 6 ( a+6 ) 5 ( a-7 ) 2

#### Solution

( a7 ) 4 ( a+6 ) 3 ( a7 ) 4 ( a+6 ) 3

### Exercise 35

( m+7 ) 4 ( m-8 ) 5 ( m+7 ) 7 ( m-8 ) 2 ( m+7 ) 4 ( m-8 ) 5 ( m+7 ) 7 ( m-8 ) 2

### Exercise 36

( a+2 ) ( a-1 ) 3 ( a+1 )( a-1 ) ( a+2 ) ( a-1 ) 3 ( a+1 )( a-1 )

#### Solution

( a+2 ) ( a1 ) 2 ( a+1 ) ( a+2 ) ( a1 ) 2 ( a+1 )

### Exercise 37

( b+6 ) ( b-2 ) 4 ( b-1 )( b-2 ) ( b+6 ) ( b-2 ) 4 ( b-1 )( b-2 )

### Exercise 38

8 ( x+2 ) 3 ( x-5 ) 6 2( x+2 ) ( x-5 ) 2 8 ( x+2 ) 3 ( x-5 ) 6 2( x+2 ) ( x-5 ) 2

#### Solution

4 ( x+2 ) 2 ( x5 ) 4 4 ( x+2 ) 2 ( x5 ) 4

### Exercise 39

14 ( x-4 ) 3 ( x-10 ) 6 -7 ( x-4 ) 2 ( x-10 ) 2 14 ( x-4 ) 3 ( x-10 ) 6 -7 ( x-4 ) 2 ( x-10 ) 2

### Exercise 40

x 2 +x-12 x 2 -4x+3 x 2 +x-12 x 2 -4x+3

#### Solution

( x+4 ) ( x1 ) ( x+4 ) ( x1 )

### Exercise 41

x 2 +3x-10 x 2 +2x-15 x 2 +3x-10 x 2 +2x-15

### Exercise 42

x 2 -10x+21 x 2 -6x-7 x 2 -10x+21 x 2 -6x-7

#### Solution

( x3 ) ( x+1 ) ( x3 ) ( x+1 )

### Exercise 43

x 2 +10x+24 x 2 +6x x 2 +10x+24 x 2 +6x

### Exercise 44

x 2 +9x+14 x 2 +7x x 2 +9x+14 x 2 +7x

#### Solution

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

### Exercise 45

6 b 2 -b 6 b 2 +11b-2 6 b 2 -b 6 b 2 +11b-2

### Exercise 46

3 b 2 +10b+3 3 b 2 +7b+2 3 b 2 +10b+3 3 b 2 +7b+2

b+3 b+2 b+3 b+2

### Exercise 47

4 b 2 -1 2 b 2 +5b-3 4 b 2 -1 2 b 2 +5b-3

### Exercise 48

16 a 2 -9 4 a 2 -a-3 16 a 2 -9 4 a 2 -a-3

#### Solution

( 4a3 ) ( a1 ) ( 4a3 ) ( a1 )

### Exercise 49

20 x 2 +28xy+9 y 2 4 x 2 +4xy+ y 2 20 x 2 +28xy+9 y 2 4 x 2 +4xy+ y 2

For the following problems, reduce each rational expression if possible. If not possible, state the answer in lowest terms.

### Exercise 50

x+3 x+4 x+3 x+4

#### Solution

( x+3 ) ( x+4 ) ( x+3 ) ( x+4 )

a+7 a-1 a+7 a-1

3a+6 3 3a+6 3

a+2 a+2

4x+12 4 4x+12 4

### Exercise 54

5a-5 -5 5a-5 -5

#### Solution

( a1 ) or a+1 ( a1 ) or a+1

6b-6 -3 6b-6 -3

### Exercise 56

8x-16 -4 8x-16 -4

2( x2 ) 2( x2 )

4x-7 -7 4x-7 -7

### Exercise 58

-3x+10 10 -3x+10 10

#### Solution

3x+10 10 3x+10 10

x-2 2-x x-2 2-x

a-3 3-a a-3 3-a

1 1

### Exercise 61

x 3 -x x x 3 -x x

### Exercise 62

y 4 -y y y 4 -y y

y 3 1 y 3 1

### Exercise 63

a 5 - a 2 a a 5 - a 2 a

### Exercise 64

a 6 - a 4 a 3 a 6 - a 4 a 3

#### Solution

a( a+1 )( a1 ) a( a+1 )( a1 )

### Exercise 65

4 b 2 +3b b 4 b 2 +3b b

### Exercise 66

2 a 3 +5a a 2 a 3 +5a a

#### Solution

2 a 2 +5 2 a 2 +5

### Exercise 67

a a 3 +a a a 3 +a

### Exercise 68

x 4 x 5 -3x x 4 x 5 -3x

#### Solution

x 3 x 4 3 x 3 x 4 3

### Exercise 69

-a - a 2 -a -a - a 2 -a

## Excercises For Review

### Exercise 70

((Reference)) Write ( 4 4 a 8 b 10 4 2 a 6 b 2 ) 1 ( 4 4 a 8 b 10 4 2 a 6 b 2 ) 1 so that only positive exponents appear.

#### Solution

1 16 a 2 b 8 1 16 a 2 b 8

### Exercise 71

((Reference)) Factor y 4 16 y 4 16 .

### Exercise 72

((Reference)) Factor 10 x 2 17x+3 10 x 2 17x+3 .

#### Solution

( 5x1 )( 2x3 ) ( 5x1 )( 2x3 )

### Exercise 73

((Reference)) Supply the missing word. An equation expressed in the form ax+by=c ax+by=c is said to be expressed in


form.

### Exercise 74

((Reference)) Find the domain of the rational expression 2 x 2 3x18 2 x 2 3x18 .

x3,6 x3,6

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