Find an nth degree polynomial function satisfying the given conditions

Question

n=3

4 & 4i are zeros

f(2)=-120

Answer 1

4 is a zero (x-4) is a factor

4i is a zero , therefore -4i is a zero which follows that (x-4i) and (x+4i) are factors

Our polynomial could be A(x-4)(x+4i)(x-4i)

(x+4i)(x-4i) simplifies tp (x^2+16)

So we have A(x-4)(x^2+16) = A(x^3-4x^2+16x-64)

You ask why do we have a constant out front?

We also must make sure the function passes through the ordered pair (2 , -120)

If we evaluate our current function at 2, we get -40A

Therefore A needs to be set at 3 to pass through -120

So our 3rd degree polynomial is 3(x^3-4x^2+16x-64) = 3x^3-12x^2+48x - 192