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.