Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 6 - Section 6.5 - A General Factoring Strategy - Exercise Set - Page 463: 107



Work Step by Step

$ 9ax^{3}+15ax^{2}-14ax\qquad$...factor out the common term, $ax$ $=ax(9x^{2}+15x-14)$ ... Searching for two factors of $ac=-126$ whose sum is $b=15,$ we find$\qquad-6$ and $21.$ Rewrite the middle term and factor in pairs: $=ax(9x^{2}-6x+21x-14)=$ $=ax[3x(3x-2)+7(3x-2)]$ =$ax(3x-2)(3x+7)$
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