Answer
$x^6-12x^5+55x^4-120x^3+139x^2-108x+85$
Work Step by Step
The Conjugate Pairs Theorem says that if a polynomial has real coefficients, then any complex zeros occur in conjugate pairs. That is, if $a + bi$ is a zero then so is $a – bi$ and vice-versa.
Therefore the zeros of the function are:
$2\pm i$, $4\pm i$ and $\pm i$
Hence, the function could be (by setting $a=1$):
$[x-(2+i)][x-(2-i)](x-i)[x-(-i)][x-(4-i)][x-(4+i)]\\
=(x-2-i)(x-2+i)(x-i)(x+i)(x-4+i)(x-4-i)\\
=x^6-12x^5+55x^4-120x^3+139x^2-108x+85$
