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**Special Types of Factoring**

Section 5.5 Special Types of Factoring

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**Objectives Difference of Two Squares Perfect Square Trinomials**

Sum and Difference of Two Cubes

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**DIFFERENCE OF TWO SQUARES**

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

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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)

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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)

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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)

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**PERFECT SQUARE TRINOMIALS**

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

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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

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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

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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.

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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

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**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)

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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)

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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)

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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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Difference of Squares Recall that, when multiplying conjugate binomials, the product is a difference of squares. E.g., (x - 7)(x + 7) = x 2 - 49 Therefore,

Difference of Squares Recall that, when multiplying conjugate binomials, the product is a difference of squares. E.g., (x - 7)(x + 7) = x 2 - 49 Therefore,

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