Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.6 Factoring: A General Strategy - 5.6 Exercise Set - Page 345: 98



Work Step by Step

Rearranging the terms, the given expression is equivalent to \begin{array}{l}\require{cancel} 18+a^3-9a-2a^2 \\\\= a^3-2a^2-9a+18 .\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} a^3-2a^2-9a+18 \\\\= (a^3-2a^2)-(9a-18) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} a^2(a-2)-9(a-2) .\end{array} Factoring the $GCF= (a-2) $ of the entire expression above results to \begin{array}{l}\require{cancel} (a-2)(a^2-9) .\end{array} The expressions $ a^2 $ and $ 9 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ a^2-9 ,$ 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^2-9) \\\\= (a-2)[(a)^2-(3)^2] \\\\= (a-2)(a-3)(a+3) .\end{array}
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