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: 41

Answer

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

Work Step by Step

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