Question

Find a polynomial function with real coefficients that has the given zeros.
-4, -4, 1+root3 i

Answer 1

By the way, if 1 + root3 i is a root, 1 - root3 i will also have to be a root. Imaginary roots ALWAYS come in conjugate pairs.
Also, did you mean to write -4 twice??
If those are the roots, then when we set the function = to 0, those numbers when plugged in must make the function = 0.
Thus, f(x) = (x - (-4))(x - (1 + root3 i))(x - (1 - root3 i))

Answer 2

in the answer key it says that it's f(x)= x^4 +6x^3 +4x^2 +64
so would it be
(x+4)(x+4)((x - ((1 + root3 i)(x - (1 - root3 i))

Answer 3

oh! ok i know what your 2 -4's imply.
Should be
(x - (-4))(x - (-4)) (x - (1 + root3 i))(x - (1 - root3 i))
Multiply that ugly thing out and I bet you get
x^4 +6x^3 +4x^2 +64

Answer 4

lol thank you