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: 38



Work Step by Step

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 $9\times 8=72$ that add to $-18$ are $-12$ and $-6.$ $9y^{2}-18y+8=9y^{2}-6y-12y+8$ $=\left(9y^{2}-6y\right)+(-12y+8)$ $=3y(3y-2)-4(3y-2)$ $=(3y-2)(3y-4)$
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.