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: 69

Answer

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

Work Step by Step

Factor out the GCF of $2$ to obtain: $=2(x^4-81)$ The binomial above can be written as $(x^2)^2-9^2$, so the expression above is equivalent to: $=2[(x^2)^2-9^2]$ Factor the difference of two squares using the formula $a^2-b^2=(a+b)(a-b)$ to obtain: $=2(x^2+9)(x^2-9) \\=2(x^2+9)(x^2-3^2)$ Factor the difference of two squares using the same formula to obtain: $=2(x^2+9)(x+3)(x-3)$
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.