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: 69

Answer

$4(a-b^2)(a^2+2b)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Factor the $GCF$ of the terms. Then, group the terms of the given expression, $ 4a^3-4a^2b^2+8ab-8b^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 $GCF= 4 ,$, the expression above is equivalent to \begin{array}{l}\require{cancel} 4\left(\dfrac{4a^3}{4}-\dfrac{4a^2b^2}{4}+\dfrac{8ab}{4}-\dfrac{8b^3}{4}\right) \\\\= 4\left(a^3-a^2b^2+2ab-2b^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} 4[(a^3-a^2b^2)+(2ab-2b^3)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 4[a^2(a-b^2)+2b(a-b^2)] .\end{array} Factoring the $GCF= (a-b^2) $ of the entire expression above results to \begin{array}{l}\require{cancel} 4[(a-b^2)(a^2+2b)] \\\\= 4(a-b^2)(a^2+2b) .\end{array}
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