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: 27

Answer

$2(d^{2}+4)(2d-3)$

Work Step by Step

$4d^{3}=2.2.d.d.d$ and $6d^{2}=2.3.d.d$ Hence, $GCF=$$2.d.d=2d^{2}$ $16d=2.2.2.2.d$ and $24=2.2.2.3$ Hence, $GCF=2.2.2=8$ After grouping: $=(4d^{3}-6d^{2})+ (16d-24)$ $=2d^{2}(2d-3)+8(2d-3)$ $=(2d^{2}+8)(2d-3)$ $=2(d^{2}+4)(2d-3)$
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