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 462: 93



Work Step by Step

$ 6a^{2}b+ab-2b\qquad$... factor out the common term, $b$ $=b(6a^{2}+a-2)$ ... Searching for two factors of $ac=-12$ whose sum is $b=1,$ we find$\qquad-3$ and $4.$ Rewrite the middle term and factor in pairs: $=b(6a^{2}-3a+4a-2)=$ $=b[3a(2a-1)+2(2a-1)]$ = $b(2a-1)(3a+2)$
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