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OBJECTIVES:

1. Factor a difference of squares. 2. Factor a perfect square trinomial. 3. Factor a difference of cubes. 4. Factor a sum of cubes.

1. Factor a difference of squares. 2. Factor a perfect square trinomial. 3. Factor a difference of cubes. 4. Factor a sum of cubes.

1.
Copyright © 2010 Pearson Education, Inc. All rights reserved

Sec 7.3 - 1

Sec 7.3 - 1

2.
Chapter 7

Factoring

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Sec 7.3 - 2

Factoring

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Sec 7.3 - 2

3.
7.3

Special Factoring

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Sec 7.3 - 3

Special Factoring

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Sec 7.3 - 3

4.
7.3 Special Factoring

Objectives

1. Factor a difference of squares.

2. Factor a perfect square trinomial.

3. Factor a difference of cubes.

4. Factor a sum of cubes.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 4

Objectives

1. Factor a difference of squares.

2. Factor a perfect square trinomial.

3. Factor a difference of cubes.

4. Factor a sum of cubes.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 4

5.
7.3 Special Factoring

The Difference of Squares

Difference of Squares

x2 – y2 = (x + y)(x – y)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 5

The Difference of Squares

Difference of Squares

x2 – y2 = (x + y)(x – y)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 5

6.
7.3 Special Factoring

EXAMPLE 1 Factoring Differences of Squares

Factor each polynomial.

(a) 2n2 – 50

There is a common factor of 2.

2n2 – 50 = 2(n2 – 25) Factor out the common factor.

= 2(n + 5)(n – 5) Factor the difference of squares.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 6

EXAMPLE 1 Factoring Differences of Squares

Factor each polynomial.

(a) 2n2 – 50

There is a common factor of 2.

2n2 – 50 = 2(n2 – 25) Factor out the common factor.

= 2(n + 5)(n – 5) Factor the difference of squares.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 6

7.
7.3 Special Factoring

EXAMPLE 1 Factoring Differences of Squares

Factor each polynomial.

(b) 9g2 – 16

9g2 – 16 = (3g)2 – (4)2 = (3g + 4)(3g – 4)

A2 – B2 (A + B)(A – B)

(c) 4h2 – (w + 5)2

4h2 – (w + 5)2 = (2h)2 – (w + 5)2 = (2h + w + 5)(2h – [w + 5])

= (2h + w + 5)(2h – w – 5)

A2 – B2 (A + B) (A – B)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 7

EXAMPLE 1 Factoring Differences of Squares

Factor each polynomial.

(b) 9g2 – 16

9g2 – 16 = (3g)2 – (4)2 = (3g + 4)(3g – 4)

A2 – B2 (A + B)(A – B)

(c) 4h2 – (w + 5)2

4h2 – (w + 5)2 = (2h)2 – (w + 5)2 = (2h + w + 5)(2h – [w + 5])

= (2h + w + 5)(2h – w – 5)

A2 – B2 (A + B) (A – B)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 7

8.
7.3 Special Factoring

Caution

Assuming no greatest common factor except 1, it is not possible to

factor (with real numbers) a sum of squares, such as x2 + 16.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 8

Caution

Assuming no greatest common factor except 1, it is not possible to

factor (with real numbers) a sum of squares, such as x2 + 16.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 8

9.
7.3 Special Factoring

Perfect Square Trinomial

Perfect Square Trinomial

x2 + 2xy + y2 = (x + y)2

x2 – 2xy + y2 = (x – y)2

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 9

Perfect Square Trinomial

Perfect Square Trinomial

x2 + 2xy + y2 = (x + y)2

x2 – 2xy + y2 = (x – y)2

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 9

10.
7.3 Special Factoring

EXAMPLE 2 Factoring Perfect Square Trinomials

Factor each polynomial.

(a) 9g2 – 42g + 49

Here 9g2 = (3g)2 and 49 = 72. The sign of the middle term is –, so if

9g2 – 42g + 49 is a perfect square trinomial, the factored form will

have to be

(3g – 7)2.

Take twice the product of the two terms to see if this is correct.

2(3g)(–7) = –42g

This is the middle term of the given trinomial, so

9g2 – 42g + 49 = (3g – 7)2.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 10

EXAMPLE 2 Factoring Perfect Square Trinomials

Factor each polynomial.

(a) 9g2 – 42g + 49

Here 9g2 = (3g)2 and 49 = 72. The sign of the middle term is –, so if

9g2 – 42g + 49 is a perfect square trinomial, the factored form will

have to be

(3g – 7)2.

Take twice the product of the two terms to see if this is correct.

2(3g)(–7) = –42g

This is the middle term of the given trinomial, so

9g2 – 42g + 49 = (3g – 7)2.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 10

11.
7.3 Special Factoring

EXAMPLE 2 Factoring Perfect Square Trinomials

Factor each polynomial.

(b) 25x2 + 60xy + 64y2

If this is a perfect square trinomial, it will equal (5x + 8y)2. By the

pattern described earlier, if multiplied out, this squared binomial has a

middle term of 2(5x)(8y), which does not equal 60xy. Verify that this

trinomial cannot be factored by the methods of the previous section

either. It is prime.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 11

EXAMPLE 2 Factoring Perfect Square Trinomials

Factor each polynomial.

(b) 25x2 + 60xy + 64y2

If this is a perfect square trinomial, it will equal (5x + 8y)2. By the

pattern described earlier, if multiplied out, this squared binomial has a

middle term of 2(5x)(8y), which does not equal 60xy. Verify that this

trinomial cannot be factored by the methods of the previous section

either. It is prime.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 11

12.
7.3 Special Factoring

EXAMPLE 2 Factoring Perfect Square Trinomials

Factor each polynomial.

(c) (n – 4)2 + 18(n – 4) + 81 = [ (n – 4) + 9 ]2

= (n + 5)2,

since 2(n – 4)9 = 18(n – 4), the middle term.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 12

EXAMPLE 2 Factoring Perfect Square Trinomials

Factor each polynomial.

(c) (n – 4)2 + 18(n – 4) + 81 = [ (n – 4) + 9 ]2

= (n + 5)2,

since 2(n – 4)9 = 18(n – 4), the middle term.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 12

13.
7.3 Special Factoring

EXAMPLE 2 Factoring Perfect Square Trinomials

Factor each polynomial.

(d) c2 – 6c + 9 – h2

Since there are four terms, we will use factoring by grouping. The first

three terms here form a perfect square trinomial. Group them together,

and factor as follows.

(c2 – 6c + 9) – h2 = (c – 3)2 – h2

The result is the difference of squares. Factor again to get

= (c – 3 + h)(c – 3 – h).

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 13

EXAMPLE 2 Factoring Perfect Square Trinomials

Factor each polynomial.

(d) c2 – 6c + 9 – h2

Since there are four terms, we will use factoring by grouping. The first

three terms here form a perfect square trinomial. Group them together,

and factor as follows.

(c2 – 6c + 9) – h2 = (c – 3)2 – h2

The result is the difference of squares. Factor again to get

= (c – 3 + h)(c – 3 – h).

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 13

14.
7.3 Special Factoring

Difference of Cubes

Difference of Cubes

x3 – y3 = (x – y)(x2 + xy + y2)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 14

Difference of Cubes

Difference of Cubes

x3 – y3 = (x – y)(x2 + xy + y2)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 14

15.
7.3 Special Factoring

EXAMPLE 3 Factoring Difference of Cubes

Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2).

(a) a3 – 125 = a3 – 53

= (a – 5)(a2 + 5a + 52)

= (a – 5)(a2 + 5a + 25)

a3 –125

Check: = (a – 5)(a2 + 5a + 25)

–5a

Opposite of the product of the cube

roots gives the middle term.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 15

EXAMPLE 3 Factoring Difference of Cubes

Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2).

(a) a3 – 125 = a3 – 53

= (a – 5)(a2 + 5a + 52)

= (a – 5)(a2 + 5a + 25)

a3 –125

Check: = (a – 5)(a2 + 5a + 25)

–5a

Opposite of the product of the cube

roots gives the middle term.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 15

16.
7.3 Special Factoring

EXAMPLE 3 Factoring Difference of Cubes

Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2).

(b) 8g3 – h3 = (2g)3 – h3

= (2g – h) [ (2g)2 + (2g)(h) + h2) ]

= (2g – h)(4g2 + 2gh + h2)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 16

EXAMPLE 3 Factoring Difference of Cubes

Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2).

(b) 8g3 – h3 = (2g)3 – h3

= (2g – h) [ (2g)2 + (2g)(h) + h2) ]

= (2g – h)(4g2 + 2gh + h2)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 16

17.
7.3 Special Factoring

EXAMPLE 3 Factoring Difference of Cubes

Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2).

(c) 64m3 – 27n3 = (4m)3 – (3n)3

= (4m – 3n) [ (4m)2 + (4m)(3n) + (3n)2 ]

= (4m – 3n)(16m2 + 12mn + 9n2)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 17

EXAMPLE 3 Factoring Difference of Cubes

Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2).

(c) 64m3 – 27n3 = (4m)3 – (3n)3

= (4m – 3n) [ (4m)2 + (4m)(3n) + (3n)2 ]

= (4m – 3n)(16m2 + 12mn + 9n2)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 17

18.
7.3 Special Factoring

Sum of Cubes

Sum of Cubes

x3 + y3 = (x + y)(x2 – xy + y2)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 18

Sum of Cubes

Sum of Cubes

x3 + y3 = (x + y)(x2 – xy + y2)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 18

19.
7.3 Special Factoring

Note on Signs

The sign of the second term in the binomial factor of a sum or difference

of cubes is always the same as the sign in the original polynomial.

In the trinomial factor, the first and last terms are always positive;

the sign of the middle term is the opposite of the sign of the second term

in the binomial factor.

Difference of Cubes x3 – y3 = (x – y)(x2 + xy + y2)

Sum of Cubes x3 + y3 = (x + y)(x2 – xy + y2)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 19

Note on Signs

The sign of the second term in the binomial factor of a sum or difference

of cubes is always the same as the sign in the original polynomial.

In the trinomial factor, the first and last terms are always positive;

the sign of the middle term is the opposite of the sign of the second term

in the binomial factor.

Difference of Cubes x3 – y3 = (x – y)(x2 + xy + y2)

Sum of Cubes x3 + y3 = (x + y)(x2 – xy + y2)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 19

20.
7.3 Special Factoring

EXAMPLE 4 Factoring Sums of Cubes

Factor each polynomial. Recall, x3 + y3 = (x + y)(x2 – xy + y2).

(a) n 3 + 8 = n3 + 23

= (n + 2)(n2 – 2n + 22)

= (n + 2)(n2 – 2n + 4)

(b) 64v3 + 27g3 = (4v)3 + (3g)3

= (4v + 3g) [ (4v)2 – (4v)(3g) + (3g)2 ]

= (4v + 3g) (16v2 – 12gv + 9g2)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 20

EXAMPLE 4 Factoring Sums of Cubes

Factor each polynomial. Recall, x3 + y3 = (x + y)(x2 – xy + y2).

(a) n 3 + 8 = n3 + 23

= (n + 2)(n2 – 2n + 22)

= (n + 2)(n2 – 2n + 4)

(b) 64v3 + 27g3 = (4v)3 + (3g)3

= (4v + 3g) [ (4v)2 – (4v)(3g) + (3g)2 ]

= (4v + 3g) (16v2 – 12gv + 9g2)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 20

21.
7.3 Special Factoring

EXAMPLE 4 Factoring Sums of Cubes

Factor each polynomial. Recall, x3 + y3 = (x + y)(x2 – xy + y2).

(c) 2k3 + 250 = 2(k3 + 125)

= 2(k3 + 53)

= 2(k + 5)(k2 – 5k + 25)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 21

EXAMPLE 4 Factoring Sums of Cubes

Factor each polynomial. Recall, x3 + y3 = (x + y)(x2 – xy + y2).

(c) 2k3 + 250 = 2(k3 + 125)

= 2(k3 + 53)

= 2(k + 5)(k2 – 5k + 25)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 21

22.
7.3 Special Factoring

Factoring Summary

Special Types of Factoring (Memorize)

Difference of Squares x2 – y2 = (x + y)(x – y)

Perfect Square Trinomial x2 + 2xy + y2 = (x + y)2

x2 – 2xy + y2 = (x – y)2

Difference of Cubes x3 – y3 = (x – y)(x2 + xy + y2)

Sum of Cubes x3 + y3 = (x + y)(x2 – xy + y2)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 22

Factoring Summary

Special Types of Factoring (Memorize)

Difference of Squares x2 – y2 = (x + y)(x – y)

Perfect Square Trinomial x2 + 2xy + y2 = (x + y)2

x2 – 2xy + y2 = (x – y)2

Difference of Cubes x3 – y3 = (x – y)(x2 + xy + y2)

Sum of Cubes x3 + y3 = (x + y)(x2 – xy + y2)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 22