#### 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.