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