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 :)