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: 71

Answer

$y^2(2x+1)(x^2-7)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Factor the $GCF$ of the terms. Then, group the terms of the given expression, $ 2x^3y^2+x^2y^2-14xy^2-7y^2 ,$ 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 $GCF= y^2 ,$, the expression above is equivalent to \begin{array}{l}\require{cancel} y^2\left( \dfrac{2x^3y^2}{y^2}+\dfrac{x^2y^2}{y^2}-\dfrac{14xy^2}{y^2}-\dfrac{7y^2}{y^2} \right) \\\\= y^2\left( 2x^3+x^2-14x-7 \right) .\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} y^2[(2x^3+x^2)-(14x+7)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} y^2[x^2(2x+1)-7(2x+1)] .\end{array} Factoring the $GCF= (2x+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} y^2[(2x+1)(x^2-7)] \\\\= y^2(2x+1)(x^2-7) .\end{array}
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