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: 50

Answer

$(a+b)(3m+2b)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Factor by grouping the first $2$ and the last $2$ terms of the given expression, $ 3ma+3mb+2ab+2b^2 .$ Then, factor the $GCF$ in each group. Finally, factor the $GCF$ of the entire expression. $\bf{\text{Solution Details:}}$ Grouping the first $2$ and the last $2$ terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (3ma+3mb)+(2ab+2b^2) .\end{array} Factoring the $GCF$ in each group, results to \begin{array}{l}\require{cancel} 3m(a+b)+2b(a+b) .\end{array} Factoring the $GCF= (a+b) $ of the entire expression above results to \begin{array}{l}\require{cancel} (a+b)(3m+2b) .\end{array}
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