A(t)=(t−k)(t−3)(t−6)(t+3) given that a(2)=0 , what is the absolute value of the product of the zeros of a?

So using a(2)=0 we can first solve for k by substituting t for 2
0 = (2-k)(2-3)(2-6)(2+3)
0 = (2-k)(-1)(-4)(5)
0 = (2-k)20
0 = 40 - 20k
-40 = -20k
k = 2

The next step would be to find all the 0s of a.
0 = (t-2)(t-3)(t-6)(t+3)
T = 2,3,6,-3

Then we find the product
2x3x6x-3 = -108

Since the problem asks for the absolute value, the answer is positive 108