Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 10 - Review - Page 726: 6

Answer

$17a^3+25a^2$

Work Step by Step

Add $11a^2$ and $14a^2$: $11a^2 + 14a^2=25a^2$ $17a^3+25a^2+a-a$ The a's cancel out, so the final result is $17a^3+25a^2$
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