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: 64

Answer

$(y^2+1)(x^3-3)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Group the terms of the given expression, $ x^3y^2-3-3y^2+x^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. $\bf{\text{Solution Details:}}$ Grouping the first and fourth terms and the second and third terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (x^3y^2+x^3)+(-3-3y^2) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x^3(y^2+1)-3(1+y^2) \\\\= x^3(y^2+1)-3(y^2+1) .\end{array} Factoring the $GCF= (y^2+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (y^2+1)(x^3-3) .\end{array}
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