Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapter 5 Test: 9



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use factoring by grouping to factor the given expression, $ a^3+2a^2-ab^2-2b^2 .$ This would result to a factor that is a difference of $2$ squares. Use then the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (a^3+2a^2)-(ab^2+2b^2) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} a^2(a+2)-b^2(a+2) .\end{array} Factoring the $GCF= (a+2) $ of the entire expression above results to \begin{array}{l}\require{cancel} (a+2)(a^2-b^2) .\end{array} The expressions $ a^2 $ and $ b^2 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ a^2-b^2 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (a+2)[(a)^2-(b^2)] \\\\= (a+2)[(a+b)(a-b)] \\\\= (a+2)(a+b)(a-b) .\end{array}
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