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Special Types of Factoring
Section 5.5 Special Types of Factoring

Objectives Difference of Two Squares Perfect Square Trinomials
Sum and Difference of Two Cubes

DIFFERENCE OF TWO SQUARES
For any real numbers a and b, a2 – b2 = (a – b)(a + b).

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

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

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)

PERFECT SQUARE TRINOMIALS
For any real numbers a and b, a2 + 2ab + b2 = (a + b)2 and a2 − 2ab + b2 = (a − b)2.

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

Example (cont) Factor. a. x2 + 8x + 16 b. 4x2 − 12x + 9 Solution b. 4x2 − 12x + 9 Start by writing as (2x)2 − 12x + 32 Check the middle term 2(2x)(3) = 12x, the middle term checks 4x2 − 12x + 9 = (2x – 3)2

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

Example (cont) Factor each expression. a. 9x2 + 6xy + y2 b. 16a3 + 8a2b + ab2 Solution b. 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

SUM AND DIFFERENCE OF TWO CUBES
For any real numbers a and b, a3 + b3 = (a + b)(a2 – ab + b2) and a3 − b3 = (a − b)(a2 + ab + b2)

Example Factor each polynomial. a. n b. 8x3 − 125y3 Solution a. n Because n3 = (n)3 and 27 = 33, we let a = n, b = 3, and factor. a3 + b3 = (a + b)(a2 – ab + b2) gives n = (n + 3)(n2 – n ∙ ) = (n + 3)(n2 – 3n + 9)

Example (cont) Factor each polynomial. a. n b. 8x3 − 125y3 Solution b. 8x3 − 125y3 8x3 = (2x)3 and 125y3 = (5y)3, so 8x3 − 125y3 = (2x)3 – (5y)3 a3 + b3 = (a + b)(a2 – ab + b2) gives (2x)3 – (5y)3 = (2x − 5y)(4x2 + 10xy + 25y2)

Example Factor the expression x6 + 64y3. Solution Let x3 = (x2)3 and b3 = (4y)3. a3 + b3 = (a + b)(a2 – ab + b2) implies x6 + 64y3 = (x2 + 4y)(x4 – 4x2y + 16y2).

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