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: 45

Answer

5a$b^{2}$(2ab + 1 - 3b)

Work Step by Step

We are given the polynomial 10$a^{2}$$b^{3}$ + 5a$b^{2}$ - 15a$b^{3}$. The GCF of all terms is 5a$b^{2}$, so we can factor that out from each term. 5a$b^{2}$(2ab) + 5a$b^{2}$(1) + 5a$b^{2}$(-3b) Next, we can use the distributive property. 5a$b^{2}$(2ab + 1 - 3b)
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