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 - Page 296: 100c

Answer

The values found in part (b) are the same because $h(t)=-16(t^2-14)$ is just a factored form of $h(t)=-16t^2+224$, which means that they are equivalent to each other.

Work Step by Step

$h(t)=-16(t^2-14)$ is just a factored form of $h(t)=-16t^2+224$. This means that −$-16(t^2-14)$ and $-16t^2+224$ are equivalent to each other. Thus, when you evaluate $h(t)=-16t^2+224$ and $h(t)=-16(t^2-14)$ for t=2, the two expressions will give the same value.
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