## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Section 5.1 - Greatest Common Factors and Factoring by Grouping - 5.1 Exercises - Page 330: 62

#### Answer

$(2b+7)(5a-3)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ Group the terms of the given expression, $10ab-21-6b+35a ,$ 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. $\bf{\text{Solution Details:}}$ Grouping the first and fourth terms and the second and third terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (10ab+35a)-(21+6b) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5a(2b+7)-3(7+2b) \\\\= 5a(2b+7)-3(2b+7) .\end{array} Factoring the $GCF= (2b+7)$ of the entire expression above results to \begin{array}{l}\require{cancel} (2b+7)(5a-3) .\end{array}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.