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: 104



Work Step by Step

$-10a^{4}b^{2}+15a^{3}b^{3}+25a^{2}b^{4}\qquad$...factor out the common term, $-5a^{2}b^{2}$. $=-5a^{2}b^{2}(2a^{2}-3ab-5b^{2})$ ... Searching for two factors of $ac=-10$ whose sum is $b=-3,$ we find$\qquad 2$ and $-5.$ Rewrite the middle term and factor in pairs: $=-5a^{2}b^{2}(2a^{2}+2ab-5ab-5b^{2})=$ $=-5a^{2}b^{2}[2a(a+b)-5b(a+b)]$ =$-5a^{2}b^{2}(a+b)(2a-5b)$
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