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: 49

Answer

$(5m+n)(2+k)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Factor by grouping the first $2$ and the last $2$ terms of the given expression, $ 10m+2n+5mk+nk .$ 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} (10m+2n)+(5mk+nk) .\end{array} Factoring the $GCF$ in each group, results to \begin{array}{l}\require{cancel} 2(5m+n)+k(5m+n) .\end{array} Factoring the $GCF= (5m+n) $ of the entire expression above results to \begin{array}{l}\require{cancel} (5m+n)(2+k) .\end{array}
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