Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 10

Answer

$(x+3)(x-y)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Group the terms of the given expression, $ x^2+3x-3y-xy ,$ such that the factored form of the groupings will result to a factor that is 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 second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (x^2+3x)-(3y+xy) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(x+3)-y(3+x) \\\\= x(x+3)-y(x+3) .\end{array} Factoring the $GCF= (x+3) $ of the entire expression above results to \begin{array}{l}\require{cancel} (x+3)(x-y) .\end{array}
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