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: 47

Answer

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

Work Step by Step

The given expression can be written as: $=(4x^2)^2-9^2$ 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 $a=4x^2$ and $b=9$ to obtain: $=(4x^2-9)(4x^2+9) \\=[(2x)^2-3^2](4x^2+9)$ Factor the difference of two squares with $a=2x$ and $b=3$ to obtain: $=(2x-3)(2x+3)(4x^2+9)$
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