Answer
a. $-16(t-2)(t+1)$
b.
We get the same answer.
The original polynomial and its factorization are equivalent, so the evaluations are equal for any value of t.
Work Step by Step
We factor out $-16,$
$-16\cdot t^{2}-16(-t)-16(-2)=-16(t^{2}-t-2)$
... search for two integer factors of $-2$ whose sum is $-1$...
... these are $-2$ and $+1$
= $-16(t-2)(t+1)$
b.
Evaluating the original polynomial:
$-16(2^{2})+16(2)+32=-64+32+32=0$
Evaluating the factorization:
$-16(2-2)(2+1)=-16(0)(3)=0$
We get the same answer.
The original polynomial and its factorization are equivalent, so the evaluations are equal for any value of t.
