Question

a(t) = (t - k)(t - 3)(t - 6)(t + 3) is a polynomial function of t, where k is a constant. Given that a(2) = 0, what is the absolute value of the product of the zeros of a?

Answer 1

Zeros:
t-k=0, t=k
t-3=0, t=3
t-6=0, t=6
t+3=0, t=-3
abs value of the product= abs value of (t*3*6*-3)=54t

Answer 2

If a(2) = 0, then k=2. The product of the zeros is
(2)*(3)*(6)*(-3) = -108

The absolute value of the product of zeros of a(t) is 108.