7 2 Factoring by GCF Classwork 012317 Simplify

7 -2 Factoring by GCF Classwork: 01/23/17 Simplify. 1. 2(w + 1) 2. 3 x(x 2 – 4) Find the GCF of each pair of monomials. 3. 4 h 2 and 6 h Holt Mc. Dougal Algebra 1

7 -2 Factoring by GCF Essential Questions How do you factor polynomials by using the greatest common factor? Holt Mc. Dougal Algebra 1

7 -2 Factoring by GCF Writing Math Aligning common factors can help you find the greatest common factor of two or more terms. Holt Mc. Dougal Algebra 1

7 -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. 2(x 2 – 2) Check 2(x 2 – 2) 2 x 2 – 4 Holt Mc. Dougal Algebra 1 The product is the original polynomial.

7 -2 Factoring by GCF Example: Factoring by Using the GCF 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 4 x(2 x 2 – x – 4) Check 4 x(2 x 2 – x – 4) 8 x 3 – 4 x 2 – 16 x Holt Mc. Dougal Algebra 1 The product is the original polynomials.

7 -2 Factoring by GCF Example: Factoring by Using the GCF 3 x 3 + 2 x 2 – 10 3 x 3 = 3 2 x 2 = 2 x x x Find the GCF. x x 10 = 2 5 There are no common factors other than 1. 3 x 3 + 2 x 2 – 10 Holt Mc. Dougal Algebra 1 The polynomial cannot be factored further.

7 -2 Factoring by GCF Example: Factoring Out a Common Binomial Factor A. 5(x + 2) + 3 x(x + 2)(5 + 3 x) Factor out (x + 2). B. – 2 b(b 2 + 1)+ (b 2 + 1)(– 2 b + 1) Holt Mc. Dougal Algebra 1 Factor out (b 2 + 1).

7 -2 Factoring by GCF Example: Factoring Out a Common Binomial Factor each expression. C. 4 z(z 2 – 7) + 9(2 z 3 + 1) There are no common – 7) + + 1) factors. The expression cannot be factored. Leave it the same way as the final answer 4 z(z 2 Holt Mc. Dougal Algebra 1 9(2 z 3

7 -2 Factoring by GCF You may be able to factor a polynomial by grouping. When a polynomial has four terms, you can make two groups and factor out the GCF from each group. Holt Mc. Dougal Algebra 1

7 -2 Factoring by GCF Example: Factoring by Grouping 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)(2 h 3 + 4) Factor out (3 h – 2). Check (3 h – 2)(2 h 3 + 4) = Holt Mc. Dougal Algebra 1 6 h 4 – 4 h 3 + 12 h – 8

7 -2 Factoring by GCF Lesson Quiz: Factor each polynomial. (byb GCF) 1. 16 x + 20 x 3 2. 4 m 4 – 12 m 2 + 8 m Factor each expression. (by grouping) 3. 7 k(k – 3) + 4(k – 3) 4. 3 y(2 y + 3) – 5(2 y + 3) Holt Mc. Dougal Algebra 1