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 6.4 - Factoring Special Forms - Exercise Set: 25

Answer

$(2x-3)(2x+3)(4x^2+9)$

Work Step by Step

RECALL: A difference of two square can be factored using the formula $a^2-b^2=(a-b)(a+b)$ The given binomial can be written as: $=(4x^2)^2-9^2$ The binomial above is a difference of two squares. Factor the difference of two squares using the formula above with $a=4x^2$ and $b=9$ to obtain: $=(4x^2-9)(4x^2+9)$ The first binomial factor can be written as: $=[(2x)^2-3^2](4x^2+9)$ The first binomial factor is a difference of two squares. Factor using the formula above with $a=2x$ and $b=3$ to obtain: $=(2x-3)(2x+3)(4x^2+9)$
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