Introductory Algebra for College Students (7th Edition)

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.
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.