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: 28



Work Step by Step

Given : $3y^{3}+14y^{2}+8y$ Now, $3.8=$$2.12$ Therefore, break the $14y^{2}$ term as $2y^{2}+12y^{2}$ This becomes : $3y^{3}+12y^{2}+2y^{2}+8y$ $3y^{3}=3.y.y.y$ and $12y^{2}=2.2.3.y.y$ Hence, $GCF=$$3.y.y=3y^{2}$ $2y^{2}=2.y.y$ and $8y=2.2.2.y$ Hence, $GCF=$$2.y=2y$ After grouping: $=(3y^{3}+12y^{2})+(2y^{2}+8y)$ $=3y^{2}(y+4)+2y(y+4)$ $=(3y^{2}+2y)(y+4)$ $=(y)(3y+2)(y+4)$
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