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: 48

Answer

$(6a-m)(5a+m)$

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 $(30)\times(-1)=-30$ that add to $+1$ are ....$+6$ and $-5$ $30a^{2}+am-m^{2}=30a^{2}-5am+6am-m^{2}$ $=\left(30a^{2}-5am\right)+\left(6am-m^{2}\right)$ $=5a(6a-m)+m(6a-m)$ $=(6a-m)(5a+m)$
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