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: 70

Answer

$5(x+3y^2)(x^2-y)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Factor the $GCF$ of the terms. Then, group the terms of the given expression, $ 5x^3+15x^2y^2-5xy-15y^3 ,$ 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 $GCF= 5 ,$, the expression above is equivalent to \begin{array}{l}\require{cancel} 5\left( \dfrac{5x^3}{5}+\dfrac{15x^2y^2}{5}-\dfrac{5xy}{5}-\dfrac{15y^3}{5} \right) \\\\= 5\left( x^3+3x^2y^2-xy-3y^3 \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} 5[(x^3+3x^2y^2)-(xy+3y^3)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5[x^2(x+3y^2)-y(x+3y^2)] .\end{array} Factoring the $GCF= (x+3y^2) $ of the entire expression above results to \begin{array}{l}\require{cancel} 5[(x+3y^2)(x^2-y)] \\\\= 5(x+3y^2)(x^2-y) .\end{array}
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