Finding the Formula for a Polynomial with imaginary roots

I'm not sure I'm setting up the problem correctly. Can you please tell me if it's right? if it's not what's the correct set-up? Thanks.

Find an nth-degree polynomial function with real coefficients satisfying the given conditions.
n=3; -5 and 4+3i are zeros; f(2)=91

f(x)=a(x-6)(x-4+3i)(x-4-3i)

(x-4)^2 = -9 for the complex roots
x^2-8x+16 = -9
(x^2-8x+25) = 0
x = -5 for the real root
(x+5) = 0
f(x) = a(x+5)(x^2-8x+25)
91 = a*7*13
a = 1
f(x) = x^3-8x^2+25x+5x^2-40x+125
f(x) = x^3-3x^2-15x+125

Nope. One of the terms should be (x+5). The complex terms look good to me, however.