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 - Section 5.5 - Factoring Special Forms - Exercise Set: 42

Answer

$(1-3x)(1+3x)(1+9x^2)$

Work Step by Step

The given binomial can be written as: $=1^2-(9x^2)^2$ The binomial is a difference of two squares. RECALL: $a^2-b^2=(a-b)(a+b)$ Factor the difference of two squares using the formula above with $a=1$ and $b=9x^2$ to obtain: $=(1-9x^2)(1+9x^2) \\=[1^2-(3x)^2)](1+9x^2)$ The first factor is a difference of two squares. Factor using the formula above with $a=1$ and $b=3x$ to obtain: $=(1-3x)(1+3x)(1+9x^2)$
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