Question

A(t)=(t−k)(t−3)(t−6)(t+3)a, left parenthesis, t, right parenthesis, equals, left parenthesis, t, minus, k, right parenthesis, left parenthesis, t, minus, 3, right parenthesis, left parenthesis, t, minus, 6, right parenthesis, left parenthesis, t, plus, 3, right parenthesis is a polynomial function of ttt, where kkk is a constant. given that a(2)=0a(2)=0a, left parenthesis, 2, right parenthesis, equals, 0, what is the absolute value of the product of the zeros of aaa?

Answer 1

Answer:

The absolute value of the product of the zeros of A(t) is 108

Step-by-step explanation:

We are given

$A(t)=(t-k)(t-3)(t-6)(t+3)$

we have

A(2)=0

we can use it and find k

we can plug t=2 , A(t)=0

$0=(2-k)(2-3)(2-6)(2+3)$

so, we get

$0=(2-k)$

$k=2$

now, we can plug it back

$A(t)=(t-2)(t-3)(t-6)(t+3)$

now, we can find zeros

$A(t)=(t-2)(t-3)(t-6)(t+3)=0$

$(t-2)=0$

$t=2$

$(t-3)=0$

$t=3$

$(t-6)=0$

$t=6$

$(t+3)=0$

$t=-3$

so, zeros are

$t=-3,t=2,t=3,t=6$

now, we can find it's product

$=-3\times 2\times 3\times 6$

$=-108$

now, we can find it's absolute value

$=|-108|$

$=108$

Answer 2

Answer:

its 11

Step-by-step explanation: