Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Chapter 6 Section 4

6.4

Special Factoring Techniques

4 1 2 3

Factor a difference of squares.

Factor a perfect square trinomial.

Factor a difference of cubes.

Factor a sum of cubes.

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Special Factoring Techniques

By reversing the rules for multiplication of binomials from

Section 5.6

, we get rules for factoring polynomials in certain forms.

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Slide 6.4 - 3

Objective

1

Factor a difference of squares.

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Slide 6.4 - 4

Factor a difference of squares.

The formula for the product of the sum and difference of the same two terms is 

x



y



x



y

 

x

2 

y

2 .

Reversing this rule leads to the following special factoring rule.

x

2 

y

2  

x



y



x



y



m

2  16 

m

2  4 2  

m

 4 

m

 4  The following conditions must be true for a binomial to be a difference of squares: 1.

2.

Both terms of the binomial must be squares, such as

x

2

,

9

y

2 , 25 , 1 ,

m

4 .

The second terms of the binomials must have different signs (one positive and one negative).

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Slide 6.4 - 5

EXAMPLE 1

Factoring Differences of Squares

Factor each binomial if possible.

t

2  81

r

2 

s

2 Solution:  9  

t

  9

s

 

x

2  16 25

t

2  10

t

2  36 

x

 4 5 prime prime

x

 4 5

After any common factor is removed, a sum of squares cannot be factored.

Slide 6.4 - 6

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 2

Factoring Differences of Squares

Factor each difference of squares.

Solution: 49

x

2  25   7

x



x

 5  64

a

2  81

b

2   8

a

 9

b

 8

a

 9

b

 You should always check a factored form by multiplying. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Slide 6.4 - 7

EXAMPLE 3

Factoring More Complex Differences of Squares

Factor completely.

50

r

2  32

z

4  100

z

4  81 Solution: 

r

2  16   

z

2  10 

z

2  10   

z

2  9 

z

2  9   

r

 

r

 4   

z

2  9  

z

 3 

z

 3 

Factor again when any of the factors is a difference of squares as in the last problem.

Check by multiplying.

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Slide 6.4 - 8

Objective

2

Factor a perfect square trinomial.

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Slide 6.4 - 9

Factor a perfect square trinomial.

The expressions 144, 4

x

2 , and 81

m

6 because 144  12 2 , 4

x

2    2 , are called

perfect squares

and 81

m

6    2 .

A

perfect square trinomial

is a trinomial that is the square of a binomial. A necessary condition for a trinomial to be a perfect square is that

two of its terms be perfect squares.

Even if two of the terms are perfect squares, the trinomial may not be a perfect square trinomial. We can multiply to see that the square of a binomial gives one of the following perfect square trinomials.

x

2  2

xy



y

2

x

2  2

xy



y

2  

x



y

 2  

x



y

 2

Slide 6.4 - 10

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EXAMPLE 4

Factoring a Perfect Square Trinomial

Factor

k

2 + 20

k

+ 100.

Solution:  

k

2 

k

 20

k

 10  2  100 Check : 2

k

10  20

k

Slide 6.4 - 11

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EXAMPLE 5

Factoring Perfect Square Trinomials

Factor each trinomial.

x

2  24

x

 144 Solution:  

x

 12  2 25

x

2  30

x

 9   5

x

 3  2 36

a

2  20

a

 25 18

x

3  84

x

2  98

x

prime  2

x

 9

x

2  42

x

 49   2

x

 3

x

 7  2

Slide 6.4 - 12

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Factoring Perfect Square Trinomials

1.

The sign of the second term in the squared binomial is always the same as the sign of the middle term in the trinomial.

2.

The first and last terms of a perfect square trinomial must be

positive

, because they are squares. For example, the polynomial

x

2 – 2

x

– 1 cannot be a perfect square, because the last term is negative.

3.

Perfect square trinomials can also be factored by using grouping or the

FOIL

method, although using the method of this section is often easier.

Slide 6.4 - 13

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Objective

3

Factor a difference of cubes.

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Slide 6.4 - 14

Factor a difference of cubes.

Just as we factored the difference of squares in Objective 1, we can also factor the

difference of cubes

by using the following pattern.

positive

x

3 

y

3  

x



y

 

x

2 

xy



y

2 

same sign opposite sign

This pattern for factoring a difference of cubes should be memorized.

The polynomial

x

3

− y

3 is not equivalent to (

x − y

) 3 , because (

x − y

) 3 can also be written as 

x

 but

x

3 

y

 

y

3    

x x x

  

y y

 

y

  

x x x

2 2   

y

 2

xy xy x

  

y

2

y



y

2  

Slide 6.4 - 15

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EXAMPLE 6

Factoring Differences of Cubes

Factor each polynomial.

x

3  216 27

x

3  8 5

x

3  5 64

x

3  125

y

6 Solution:  

x

 6    3

x

  

x

2 

x

2 6 

x

 6

x

36   4     5  4

x x

3   1  5

y

2   16 5 

x

2

x

  1  

x

2 20

xy

2  25  1  A common error in factoring a difference of cubes, such as

x

3

− y

3

=

(

x − y

)(

x

2

+ xy + y

2 ) , is to try to factor

x

2

+ xy

+

y

2 . It is easy to confuse this factor with the perfect square trinomial

x

2 + 2

xy

+

y

2 . But because there is no 2, it is unusual to be able to further factor an expression of the form

x

2

+ xy +y

2 .

Slide 6.4 - 16

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Objective

4

Factor a sum of cubes.

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Slide 6.4 - 17

Factor a sum of cubes.

A sum of squares, such as

m

2 + 25 , cannot be factored by using real numbers, but a

sum of cubes

can be factored by the following pattern.

positive

x

3 

y

3  

x



y

 

x

2 

xy



y

2 

same sign opposite sign

Note the similarities in the procedures for factoring a sum of cubes and a difference of cubes.

1.

Both are the product of a binomial and a trinomial.

2.

The binomial factor is found by remembering the “cube root, same sign, cube root.” 3.

The trinomial factor is found by considering the binomial factor and remembering, “square first term, opposite of the product, square last term.” Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Slide 6.4 - 18

The methods of factoring discussed in this section are summarized here.

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Slide 6.4 - 19

EXAMPLE 7

Factoring Sums of Cubes

Factor each polynomial.

p

3  64 Solution:  

p

 4  

p

2  4

p

 16  27

x

3  64

y

3   3

x

 4

y

  9

x

2  12

xy

 16

y

2  512

a

6 

b

3   8

a

2 

b

 64

a

4  8 2  2  Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Slide 6.4 - 20