8 2 Factoring by GCF Objective Factor polynomials
8 -2 Factoring by GCF Objective Factor polynomials by using the greatest common factor. Holt Algebra 1
8 -2 Factoring by GCF Example 1 A: Factoring by Using the GCF Factor each polynomial. Check your answer. 2 x 2 – 4 2 x 2 = 2 x x 4=2 2 Find the GCF. 2 2 x 2 – (2 2) The GCF of 2 x 2 and 4 is 2. Write terms as products using the GCF as a factor. Use the Distributive Property to factor out the GCF. Multiply to check your answer. The product is the original polynomial. 2(x 2 – 2) Check 2(x 2 – 2) 2 x 2 – 4 Holt Algebra 1
8 -2 Factoring by GCF Example 1 B: Factoring by Using the GCF Factor each polynomial. Check your answer. 8 x 3 – 4 x 2 – 16 x 8 x 3 = 2 2 2 x x x Find the GCF. 4 x 2 = 2 2 x x 16 x = 2 2 x The GCF of 8 x 3, 4 x 2, and 16 x is 4 x. 2 2 x = 4 x Write terms as products using the GCF as a factor. 2 x 2(4 x) – x(4 x) – 4(4 x) Use the Distributive Property to 4 x(2 x 2 – x – 4) factor out the GCF. Check 4 x(2 x 2 – x – 4) Multiply to check your answer. The product is the original 8 x 3 – 4 x 2 – 16 x polynomials. Holt Algebra 1
8 -2 Factoring by GCF Example 1 C: Factoring by Using the GCF Factor each polynomial. Check your answer. – 14 x – 12 x 2 – 1(14 x + 12 x 2) 14 x = 2 7 x 12 x 2 = 2 2 3 x x 2 – 1[7(2 x) + 6 x(2 x)] – 1[2 x(7 + 6 x)] – 2 x(7 + 6 x) Holt Algebra 1 Both coefficients are negative. Factor out – 1. Find the GCF. 2 The GCF of 14 x and 12 x x = 2 x is 2 x. Write each term as a product using the GCF. Use the Distributive Property to factor out the GCF.
8 -2 Factoring by GCF Example 1 D: Factoring by Using the GCF Factor each polynomial. Check your answer. 3 x 3 + 2 x 2 – 10 3 x 3 = 3 2 x 2 = 10 = x x x Find the GCF. 2 x x 2 5 3 x 3 + 2 x 2 – 10 There are no common factors other than 1. The polynomial cannot be factored further. Holt Algebra 1
8 -2 Factoring by GCF Check It Out! Example 1 a Factor each polynomial. Check your answer. 6 b + 9 b 3 5 b = 5 b 9 b = 3 3 b b b b 5(b) + 9 b 2(b) b(5 + 9 b 2) Check b(5 + 9 b 2) 5 b + 9 b 3 Holt Algebra 1 Find the GCF. The GCF of 5 b and 9 b 3 is b. Write terms as products using the GCF as a factor. Use the Distributive Property to factor out the GCF. Multiply to check your answer. The product is the original polynomial.
8 -2 Factoring by GCF Check It Out! Example 1 c Factor each polynomial. Check your answer. – 18 y 3 – 12 y 2 – 1(18 y 3 + 7 y 2) Both coefficients are negative. Factor out – 1. 18 y 3 = 2 3 3 y y y Find the GCF. 7 y 2 = 7 y y y y = y 2 The GCF of 18 y 3 and 7 y 2 is y 2. – 1[18 y(y 2) + 7(y 2)] – 1[y 2(18 y + 7)] –y 2(18 y + 7) Holt Algebra 1 Write each term as a product using the GCF. Use the Distributive Property to factor out the GCF. .
8 -2 Factoring by GCF Sometimes the GCF of terms is a binomial. This GCF is called a common binomial factor. You factor out a common binomial factor the same way you factor out a monomial factor. Holt Algebra 1
8 -2 Factoring by GCF Example 3: Factoring Out a Common Binomial Factor each expression. A. 5(x + 2) + 3 x(x + 2)(5 + 3 x) The terms have a common binomial factor of (x + 2). Factor out (x + 2). B. – 2 b(b 2 + 1)+ (b 2 + 1) – 2 b(b 2 + 1) + (b 2 + 1) The terms have a common binomial factor of (b 2 + 1). – 2 b(b 2 + 1) + 1(b 2 + 1) = 1(b 2 + 1)(– 2 b + 1) Holt Algebra 1 Factor out (b 2 + 1).
8 -2 Factoring by GCF Check It Out! Example 3 Factor each expression. c. 3 x(y + 4) – 2 y(x + 4) There are no common factors. The expression cannot be factored. d. 5 x(5 x – 2) – 2(5 x – 2) (5 x – 2)2 Holt Algebra 1 The terms have a common binomial factor of (5 x – 2 ). (5 x – 2) = (5 x – 2)2
8 -2 Factoring by GCF Example 4 A: Factoring by Grouping Factor each polynomial by grouping. Check your answer. 6 h 4 – 4 h 3 + 12 h – 8 (6 h 4 – 4 h 3) + (12 h – 8) Group terms that have a common number or variable as a factor. 2 h 3(3 h – 2) + 4(3 h – 2) Factor out the GCF of each group. 2 h 3(3 h – 2) + 4(3 h – 2) is another common factor. (3 h – 2)(2 h 3 + 4) Holt Algebra 1 Factor out (3 h – 2).
8 -2 Factoring by GCF Example 4 B: Factoring by Grouping Factor each polynomial by grouping. Check your answer. 5 y 4 – 15 y 3 + y 2 – 3 y (5 y 4 – 15 y 3) + (y 2 – 3 y) Group terms. 5 y 3(y – 3) + y(y – 3) Factor out the GCF of each group. 5 y 3(y – 3) + y(y – 3) is a common factor. (y – 3)(5 y 3 + y) Factor out (y – 3). Holt Algebra 1
8 -2 Factoring by GCF Check It Out! Example 4 a Factor each polynomial by grouping. Check your answer. 6 b 3 + 8 b 2 + 9 b + 12 (6 b 3 + 8 b 2) + (9 b + 12) Group terms. 2 b 2(3 b + 4) + 3(3 b + 4) Factor out the GCF of each group. (3 b + 4) is a common factor. 2 b 2(3 b + 4) + 3(3 b + 4)(2 b 2 + 3) Holt Algebra 1 Factor out (3 b + 4).
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