Answer
The values in part (b) are the same because the original expression and the factored expression are equivalent.
Work Step by Step
$h(t)=-16t(t-4)$ is just a factored form of $h(t)=-16t^2+64t$.
This means that $-16t(t-4)$ and $-16t^2+64t$ are equivalent to each other.
Thus, when you evaluate $h(t)=-16t^2+64t$ and $h(t)=-16t(t-4)$ for $t=1$, the two expressions will give the same value.
