## Algebra 1

Published by Prentice Hall

# Chapter 8 - Polynomials and Factoring - 8-8 Factoring by Grouping - Practice and Problem-Solving Exercises: 25

#### Answer

$3q(q+2)(q-2)(2q+1)$

#### Work Step by Step

$6q^{4}=2.3.q.q.q.q$ and $3q^{3}=3.q.q.q$ Hence, $GCF=$$3.q.q.q=3q^{3} 24q^{2}=2.2.2.3.q.q and 12q=2.2.3.q Hence, GCF=$$2.2.3.q=12q$ After grouping : $=(6q^{4}+3q^{3})-(24q^{2}+12q)$ $=3q^{3}(2q+1)-12q(2q+1)$ $=(3q^{3}-12q)(2q+1)$ $=(3q)(q^{2}-4)(2q+1)$ $=3q(q+2)(q-2)(2q+1)$ (Since $(a^{2}-b^{2})$$=(a+b)(a-b))$

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