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 - 5.2 Factoring Trinomials of the Type x2+bx+c - 5.2 Exercise Set: 50

Answer

$-a^2(a+12)(a-11)$

Work Step by Step

Factoring the negative $GCF= -a^2 $, the given expression, $ -a^4-a^3+132a^2 ,$ is equivalent to \begin{array}{l} -a^2(a^2+a-132) .\end{array} The 2 numbers whose product is $ac= 1(-132)=-132 $ and whose sum is $b= 1 $ are $\left\{ 12,-11 \right\}.$ Using these two numbers to decompose the middle term of the above trinomial, then \begin{array}{l} -a^2(a^2+12a-11a-132) \\\\= -a^2[(a^2+12a)-(11a+132)] \\\\= -a^2[a(a+12)-11(a+12)] \\\\= -a^2[(a+12)(a-11)] \\\\= -a^2(a+12)(a-11) .\end{array}
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