Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Section 5.5

Special Types of Factoring

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Objectives

• Difference of Two Squares

• Perfect Square Trinomials

• Sum and Difference of Two Cubes

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

For any real numbers a and b,

a2 – b2 = (a – b)(a + b).

DIFFERENCE OF TWO SQUARES

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Example

Factor each difference of two squares.a. 9x2 – 16 b. 5x2 + 8y2 c. 25x2 – 16y2

Solutiona. 9x2 – 16

b. Because 5x2 + 8y2 is the sum of two squares, it cannot be factored.

c. 25x2 – 16y2

= (3x)2 – (4)2 = (3x – 4) (3x + 4)

5 4 5 4x y x y

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Example

Factor each expression.a. (x + 3)2 − 16 b. 8s2 – 32t4 c. x3 + x2 − 4x − 4Solutiona. (x + 3)2 − 16

b. Factor out the common factor of 8.

= (x + 3)2 – 42 = ((x + 3) – 4))((x + 3) + 4))

= (x – 1)(x + 7)

8s2 – 32t4 = 8(s2 – 4t4)

= 8(s2 – (2t2)2)

= 8(s – 2t2)(s + 2t2)

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Example (cont)

c. x3 + x2 − 4x − 4 Start factoring by using grouping and then factor the difference of squares.

x3 + x2 − 4x − 4 = (x3 + x2) + (−4x − 4)

= x2(x + 1) − 4(x + 1)

= (x2 − 4)(x + 1)

= (x − 2)(x + 2)(x + 1)

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

For any real numbers a and b,

a2 + 2ab + b2 = (a + b)2 and

a2 − 2ab + b2 = (a − b)2.

PERFECT SQUARE TRINOMIALS

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Example

Factor. a. x2 + 8x + 16 b. 4x2 − 12x + 9

Solutiona. x2 + 8x + 16

Start by writing as x2 + 8x + 42 Check the middle term

2(x)(4) = 8x, the middle term checks

x2 + 8x + 16 = (x + 4)2

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Example (cont)

Factor.a. x2 + 8x + 16 b. 4x2 − 12x + 9

Solutionb. 4x2 − 12x + 9

Start by writing as (2x)2 − 12x + 32

Check the middle term2(2x)(3) = 12x, the middle term checks

4x2 − 12x + 9 = (2x – 3)2

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Example

Factor each expression.a. 9x2 + 6xy + y2 b. 16a3 + 8a2b + ab2

Solutiona. 9x2 + 6xy + y2

Let a2 = (3x)2 and b2 = y2.

2ab = 2(3x)(y) = 6xy, which equals the given middle term. Thus a2 + 2ab + b2 = (a + b)2 implies

9x2 + 6xy + y2 = (3x + y)2.

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Example (cont)

Factor each expression.a. 9x2 + 6xy + y2 b. 16a3 + 8a2b + ab2

Solutionb. 16a3 + 8a2b + ab2

Factor out the common factor of a.

Then factor the resulting perfect square trinomial.

16a3 + 8a2b + ab2 = a(16a2 + 8ab + b2)

= a(4a + b)2

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

For any real numbers a and b,

a3 + b3 = (a + b)(a2 – ab + b2) and

a3 − b3 = (a − b)(a2 + ab + b2)

SUM AND DIFFERENCE OF TWO CUBES

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Example

Factor each polynomial.a. n3 + 27 b. 8x3 − 125y3

Solutiona. n3 + 27Because n3 = (n)3 and 27 = 33, we let a = n, b = 3, and factor.

a3 + b3 = (a + b)(a2 – ab + b2) gives n3 + 33 = (n + 3)(n2 – n ∙ 3 + 32) = (n + 3)(n2 – 3n + 9)

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Example (cont)

Factor each polynomial.a. n3 + 27 b. 8x3 − 125y3

Solutionb. 8x3 − 125y3

8x3 = (2x)3 and 125y3 = (5y)3, so8x3 − 125y3 = (2x)3 – (5y)3

a3 + b3 = (a + b)(a2 – ab + b2) gives (2x)3 – (5y)3 = (2x − 5y)(4x2 + 10xy + 25y2)

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Example

Factor the expression x6 + 64y3.

SolutionLet x3 = (x2)3 and b3 = (4y)3.

a3 + b3 = (a + b)(a2 – ab + b2) implies

x6 + 64y3 = (x2 + 4y)(x4 – 4x2y + 16y2).