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

the absolute value of the product of the zeros of a is $-108$ .

Step-by-step explanation:

Here we have , $a(t) = (t - k)(t - 3)(t - 6)(t + 3)$ is a polynomial function of t, where k is a constant.  Given that a(2) = 0 . We need to find the  absolute value of the product of the zeros of a . Let's find out:

Equation every factor of a(t) to zero we get:

⇒ $a(t) = (t - k)(t - 3)(t - 6)(t + 3) = 0$

⇒ $(t - k) =0\\(t - 3)=0\\(t - 6)=0\\(t + 3) = 0$

⇒ $t =k\\t =3\\t = 6\\t =- 3$

But , t=2 So , $k=2$  . Now , the absolute value of the product of the zeros of a is :

⇒ $k(3)(6)(-3)$

⇒ $2(3)(6)(-3)$

⇒ $-108$

Therefore, the absolute value of the product of the zeros of a is $-108$ .