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: 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: $=(2x-3)(2x+3)(4x^2+9)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.