Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.4 Factoring Polynomials - R.4 Exercises - Page 44: 49



Work Step by Step

Factor out the gratest common factor, $2a^{2}$ $=2a^{2}\left(12a^{2}+5ab-2b^{2}\right)$ When factoring $ax^{2}+bx+c$, we search for factors of $ac$ whose sum is $b,$ and, if we find them, we rewrite $bx$ and proceed to factor in groups. Here, factors of $(12)\times(-2)=-24$ that add to $+5$ are ....$+8$ and $-3 .$ $=2a^{2}\left[ \left(12a^{2}-3ab\right)+\left(8ab-2b^{2}\right) \right]$ $=2a^{2}\left[ 3a(4a-b)+2b(4a-b) \right]$ $=2a^{2}\left[ (4a-b)(3a+2b) \right]$ $=2a^{2}(4a-b)(3a+2b)$
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.