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: 35



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 $6\times 4=24$ that add to $-11$ are $-8$ and $-3.$ $6a^{2}-11a+4=6a^{2}-3a-8a+4$ $=\left(6a^{2}-3a\right)+(-8a+4)$ $=3a(2a-1)-4(2a-1)$ $=(2a-1)(3a-4)$
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