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
