Holt Algebra Choosing a Factoring Method 8-6 Choosing a Factoring Method Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson Quiz

Holt Algebra Choosing a Factoring Method Warm Up Factor each trinomial. 1. x x x 2 – 18x – 8 3. Factor the perfect-square trinomial 16x x Factor 9x 2 – 25y 2 using the difference of two squares. (x + 5)(x + 8) (4x + 5)(4x + 5) (5x + 2)(x – 4) (3x + 5y)(3x – 5y)

Holt Algebra Choosing a Factoring Method Choose an appropriate method for factoring a polynomial. Combine methods for factoring a polynomial. Objectives

Holt Algebra Choosing a Factoring Method Recall that a polynomial is in its fully factored form when it is written as a product that cannot be factored further. GCF 4 Terms --> Grouping 3 Terms --> unfoiling 2 Terms --> Difference of Squares if a 2 - b 2

Holt Algebra Choosing a Factoring Method If none of the factoring methods work, the polynomial is said to be unfactorable. For a polynomial of the form ax 2 + bx + c, if there are no numbers whose sum is b and whose product is ac, then the polynomial is unfactorable. Helpful Hint

Holt Algebra Choosing a Factoring Method x is a sum of squares, and cannot be factored. Caution

Holt Algebra Choosing a Factoring Method Example 1A: Factoring Factor 10x x + 32 completely. Check your answer. 10x x (5x x + 16) 2(5x + 4)(x + 4) Factor out the GCF. Check 2(5x + 4)(x + 4) = 2(5x x + 4x + 16) = 10x x + 8x + 32 = 10x x + 32 Factor remaining trinomial.

Holt Algebra Choosing a Factoring Method Example 1B: Factoring Factor 8x 6 y 2 – 18x 2 y 2 completely. Check your answer. 8x 6 y 2 – 18x 2 y 2 2x 2 y 2 (4x 4 – 9) Factor out the GCF. 4x 4 – 9 is a perfect-square trinomial of the form a 2 – b 2. 2x 2 y 2 (2x 2 – 3)(2x 2 + 3) a = 2x, b = 3 Check 2x 2 y 2 (2x 2 – 3)(2x 2 + 3) = 2x 2 y 2 (4x 4 – 9) = 8x 6 y 2 – 18x 2 y 2

Holt Algebra Choosing a Factoring Method Example 1C Factor each polynomial completely. Check your answer. 4x x x 4x(x 2 + 4x + 4) 4x(x + 2) 2 Factor out the GCF. x 2 + 4x + 4 is a perfect-square trinomial of the form a 2 + 2ab + b 2. a = x, b = 2 Check 4x(x + 2) 2 = 4x(x 2 + 2x + 2x + 4) = 4x(x 2 + 4x + 4) = 4x x x

Holt Algebra Choosing a Factoring Method Example 1D Factor each polynomial completely. Check your answer. 2x 2 y – 2y 3 2y(x + y)(x – y) Factor out the GCF. 2y(x 2 – y 2 ) is a perfect-square trinomial of the form a 2 – b 2. a = x, b = y Check 2y(x + y)(x – y) = 2y(x 2 + xy – xy – y 2 ) = 2x 2 y – 2y 3 2x 2 y – 2y 3 2y(x2 – y2)2y(x2 – y2) = 2x 2 y +2xy 2 – 2xy 2 – 2y 3

Holt Algebra Choosing a Factoring Method Example 1E: Factoring Factor each polynomial completely. 9x 2 + 3x – 2 ( x + )( x + ) The GCF is 1 and there is no pattern. a = 9 and c = –2; Outer + Inner = 3 Factors of 9 Factors of 2 Outer + Inner 1 and 9 1 and –21(–2) + 1(9) = 7 3 and 3 1 and –23(–2) + 1(3) = –3 3 and 3 –1 and 23(2) + 3(–1) = 3   (3x – 1)(3x + 2)

Holt Algebra Choosing a Factoring Method Example 1F: Factoring Factor each polynomial completely. 12b b b (x + )(x + ) The GCF is 12b; (b 2 + 4b + 4) is a perfect-square trinomial in the form of a 2 + 2ab + b 2. a = 2 and c = 2 12b(b 2 + 4b + 4) 12b(b + 2)(b + 2) 12b(b + 2) 2 Factors of 4 Sum 1 and and 2 4 

Holt Algebra Choosing a Factoring Method Example 1G: Factoring Factor each polynomial completely. 4y y – 72 4(y 2 + 3y – 18) Factor out the GCF. There is no pattern. b = 3 and c = –18; look for factors of –18 whose sum is 3. (y + )(y + ) Factors of –18 Sum –1 and –2 and 9 7  –3 and 6 3  4(y – 3)(y + 6) The factors needed are –3 and 6

Holt Algebra Choosing a Factoring Method Example 1H: Factoring Factor each polynomial completely. x4 – x2x4 – x2 x 2 (x 2 – 1) Factor out the GCF. x 2 (x + 1)(x – 1) x 2 – 1 is a difference of two squares.

Holt Algebra Choosing a Factoring Method Example 1I Factor each polynomial completely. 3x 2 + 7x + 4 ( x + )( x + ) a = 3 and c = 4; Outer + Inner = 7 Factors of 3 Factors of 4 Outer + Inner 3 and 1 1 and 43(4) + 1(1) = 13 3 and 1 2 and 23(2) + 1(2) = 8 3 and 1 4 and 13(1) + 1(4) = 7   (3x + 4)(x + 1)

Holt Algebra Choosing a Factoring Method Example 1J Factor each polynomial completely. 2p p 4 – 12p 3 2p 3 (p 2 + 5p – 6) Factor out the GCF. There is no pattern. b = 5 and c = –6; look for factors of –6 whose sum is 5. (p + )(p + ) Factors of – 6 Sum – 1 and 6 5 2p 3 (p + 6)(p – 1) The factors needed are –1 and 6

Holt Algebra Choosing a Factoring Method Example 1K Factor each polynomial completely. Factor out the GCF. There is no pattern. 3q 4 (3q q + 8) 9q q q 4 a = 3 and c = 8; Outer + Inner = 10 ( q + )( q + ) Factors of 3 Factors of 8 Outer + Inner 3 and 1 1 and 83(8) + 1(1) = 25 3 and 1 2 and 43(4) + 1(2) = 14 3 and 1 4 and 23(2) + 1(4) = 10   3q 4 (3q + 4)(q + 2)

Holt Algebra Choosing a Factoring Method Example 1L Factor each polynomial completely. 2x (x 4 + 9) Factor out the GFC. x is the sum of squares and that is not factorable. 2(x 4 + 9) is completely factored.

Holt Algebra Choosing a Factoring Method

Holt Algebra Choosing a Factoring Method Tell whether the polynomial is completely factored. If not, factor it. 1. (x + 3)(5x + 10) 2. 3x 2 (x 2 + 9) Lesson Quiz (x + 4)(x 2 + 3) completely factored no; 5(x+ 3)(x + 2) 4(x + 6)(x – 2) 5. 18x 2 – 3x – x 2 – 50y x – 20x – 28x 2 3(3x + 1)(2x – 1) 2(3x + 5y)(3x – 5y) (1 + 2x)(1 – 2x)(5x + 7) Factor each polynomial completely. Check your answer. 3. x 3 + 4x 2 + 3x x x – 48