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
Option A is wrong because,
only when p(5)=0 , x-5 can be a factor of p(x)
but the question says P(3)=-2
option B is wrong because,
only when p(2)=0 , x-2 can be a factor of p(x)
but the question says P(3)=-2
option C is wrong because,
only when p(-2)=0 , x+2 can be a factor of p(x)
but the question says P(3)=-2
option D is correct because,
only when p(3)=0 , x-3 can be a factor of p(x)
here p(x) is not equal to 0
therefore, p(x)=0+(-2)=-2
so when p(x) is divided by x-3 it will leave a reminder -2
hope it helps!!
