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: 43


$(3p+2k)(-2p+3k)$ or $(2k+3p)(3k-2p)$

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 $(6)\times(-6)=-36$ that add to $+5$ are ....$+9$ and $-4.$ $6k^{2}+5kp-6p^{2}=6k^{2}-4kp+9kp-6p^{2}$ $=\left(-6p^{2}-4kp\right)+\left(9kp+6k^{2}\right)$ $=-2p(3p+2k)+3k(3p+2k)$ $=(3p+2k)(-2p+3k)$ $=(2k+3p)(3k-2p)$
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