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 R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set - Page 970: 17



Work Step by Step

Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} x^3+2x^2-x-2 \\\\= (x^3+2x^2)-(x+2) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x^2(x+2)-(x+2) .\end{array} Factoring the $GCF= (x+2) $ of the entire expression above results to \begin{array}{l}\require{cancel} (x+2)(x^2-1) .\end{array} The expressions $ x^2 $ and $ 1 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ x^2-1 ,$ 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} (x+2)[(x)^2-(1)^2] \\\\= (x+2)[(x+1)(x-1)] \\\\= (x+2)(x+1)(x-1) .\end{array}
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