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?

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

Step-by-step explanation:

We are given

we have

A(2)=0

we can use it and find k

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

so, we get

now, we can plug it back

now, we can find zeros

so, zeros are

now, we can find it's product

now, we can find it's absolute value

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb