Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Review Exercises - Page 400: 76



Work Step by Step

Given: $x^2+6x+9-4a^2$ or, $=x^2+3x+3x+9-4a^2$ This implies $=(x+3)(x+3)-(2a)^2$ or, $=(x+3)^2-(2a)^2$ Hence, Factors are: $(x+3-2a),(x+3+2a)$
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