Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.1 - Greatest Common Factors and Factoring by Grouping - 5.1 Exercises - Page 330: 68



Work Step by Step

$\bf{\text{Solution Outline:}}$ Factor the $GCF$ of the terms. Then, group the terms of the given expression, $ 12-6q-18p+9pq ,$ such that the factored form of the groupings will result to a factor common to the entire expression. Then, factor the $GCF$ in each group. Finally, factor the $GCF$ of the entire expression. $\bf{\text{Solution Details:}}$ Using the $GCF= 3 ,$, the expression above is equivalent to \begin{array}{l}\require{cancel} 3(4-2q-6p+3pq) .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} 3[(4-2q)-(6p-3pq)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3[2(2-q)-3p(2-q)] .\end{array} Factoring the $GCF= (2-q) $ of the entire expression above results to \begin{array}{l}\require{cancel} 3[(2-q)(2-3p)] \\\\= 3(2-q)(2-3p) .\end{array}
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