Algebra 1

by Hall, Prentice

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 519: 18

Answer

$(4k-9)(3k^{2}-10)$

Work Step by Step

$12k^{3}=2.2.3.k.k.k$ and $27k^{2}=3.3.3.k.k$ Hence, $GCF=$$3.k.k=3k^{2}$ $40k=2.2.2.5.k$ and $90=2.3.3.5$ Hence, $GCF=2.5=10$ After grouping: $=(12k^{3}-27k^{2})-(40k-90)$ $=3k^{2}(4k-9)-10(4k-9)$ $=(4k-9)(3k^{2}-10)$

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