Question

a(t)=(t−k)(t−3)(t−6)(t+3) is a polynomial function of t, where k is a constant. Given that a(2)=0 , what is the absolute value of the product of the zeros of a?

Answer 1

First, we need to define k because it is a missing zero;

Knowing that a(2)=0;

a(t)=(t-k)(t-3)(t-6)(t+3)
a(2)=(2-k)(2-3)(2-6)(2+3)=0
(2-k)(-1)(-4)(5)=0
(2-k)(20)=0
(2-k)=0/20
2-k=0
2=k

Therefore, the final equation is a(t)=(t-2)(t-3)(3-6)(t+3) with the zeros being (2, 3, 6, -3)

Hope I helped :)