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: 50

Answer

$3x^{3}(3x-5z)(2x+5z) $

Work Step by Step

Factor out the gratest common factor, $3x^{3}$ $=3x^{3}\left(6x^{2}+5xz-25z^{2}\right)$ 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 $(6)\times(-25)=-150$ that add to $+5$ are ....$+15$ and $-10 .$ $=3x^{3}\left[ 6x^{2}-10xz+15xz-25z^{2} \right]$ $=3x^{3}\left[ \left(6x^{2}-10xz\right)+\left(15xz-25z^{2}\right) \right]$ $=3x^{3}\left[ 2x(3x-5z)+5z(3x-5z) \right]$ $=3x^{3}\left[ (3x-5z)(2x+5z) \right]$ $=3x^{3}(3x-5z)(2x+5z) $
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