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: 99

Answer

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

Work Step by Step

Factoring the negative $GCF= -1 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} -x^4+7x^2+18 \\\\= -(x^4-7x^2-18) .\end{array} Using the factoring of trinomials in the form $x^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} (x^4-7x^2-18) \end{array} has $c= -18 $ and $b= -7 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -9,2 \right\}.$ Using these two numbers, the $\text{ expression }$ above is equivalent to \begin{array}{l}\require{cancel} -(x^2-9)(x^2+2) .\end{array} The expressions $ x^2 $ and $ 9 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ x^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} -[(x)^2-(3)^2](x^2+2) \\\\= -(x+3)(x-3)(x^2+2) .\end{array}
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