Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.5 - The Greatest Common Factor and Factoring by Grouping - Exercise Set: 40

Answer

$3b^{2}(4b^{2}+1)$

Work Step by Step

We are given the polynomial $12b^{4}+3b^{2}$. The GCF of both terms is$3b^{2}$, so we can factor $3b^{2}$ out from each term. $3b^{2}(4b^{2})+3b^{2}(1)$ Next, we can apply the distributive property. $3b^{2}(4b^{2}+1)$
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