Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.5 - Factoring Polynomials - Exercise Set - Page 68: 48



Work Step by Step

The given expression can be written as $(9x^2)^2-1^2.$ The expression above 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 $a=9x^2$ and $b=1$ to obtain $(9x^2-1)(9x^2+1) \\=[(3x)^2-1^2](9x^2+1).$ Factor the difference of two squares with $a=3x$ and $b=1$ to obtain $(3x-1)(3x+1)(9x^2+1).$
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