Answer
a. $-16(t-3)(t+1)$
b.
We get the same answer, 0.
The original polynomial and its factorization are equivalent, so the evaluations are equal for any value of t.
Work Step by Step
a.
We factor out $-16,$
$-16\cdot t^{2}-16(-2t)-16(-3)=-16(t^{2}-2t-3)$
... search for two integer factors of $-3$ whose sum is $-2$...
... these are $-3$ and $+1$
= $-16(t-3)(t+1)$
b.
Evaluating the original polynomial:
$-16(3^{2})+32(3)+48=-144+96+48=0$
Evaluating the factorization:
$-16(3-3)(3+1)=-16(0)(4)=0$
We get the same answer.
The original polynomial and its factorization are equivalent, so the evaluations are equal for any value of t.