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: 61

Answer

$6a^{8}(a+7)(a-2)$

Work Step by Step

Factoring the $GCF= 6a^8 $, the given expression, $ 6a^{10}+30a^9-84a^8 ,$ is equivalent to \begin{array}{l} 6a^{8}(a^2+5a-14) .\end{array} The 2 numbers whose product is $ac= 1(-14)=-14 $ and whose sum is $b= 5 $ are $\left\{ 7,-2 \right\}.$ Using these two numbers to decompose the middle term of the above trinomial, then \begin{array}{l} 6a^{8}(a^2+7a-2a-14) \\\\= 6a^{8}[(a^2+7a)-(2a+14)] \\\\= 6a^{8}[a(a+7)-2(a+7)] \\\\= 6a^{8}[(a+7)(a-2)] \\\\= 6a^{8}(a+7)(a-2) .\end{array}
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