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: 89



Work Step by Step

$ 3a^{2}+27ab+54b^{2}\qquad$... factor out the common term, $3$ $=3(a^{2}+9ab+18b^{2})$ ... Searching for two factors of $ac=18$ whose sum is $b=9,$ we find$\qquad 6$ and $3.$ Rewrite the middle term and factor in pairs: $=3(a^{2}+6ab+3ab+18b^{2})=$ $=3[a(a+6b)+3b(a+6b)]$ = $3(a+6b)(a+3b)$
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