Question

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

Answer 1

Since $a(2)=0$, we know that $t-2$ must be a factor of $a(t)$, so $k=2$. Then the zeros of $a(t)$ are $t=2,3,6,-3$, and their product is -108, whose absolute value is 108.