Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 8 - Polynomials and Factoring - 8-8 Factoring by Grouping - Practice and Problem-Solving Exercises - Page 520: 31



Work Step by Step

Given: $9t(t^{2}-10t+16)$ Now, $1\times16=$$-2\times-8$ Therefore, break the term $-10t$ as $(-2t-8t)$ This becomes: $=9t(t^{2}-2t-8t+16)$ $=9t(t(t-2)-8(t-2))$ $=9t((t-8)(t-2))$ $=9t(t-8)(t-2)$
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