Answer
$ x^3-14x^2+65 x -102$
Work Step by Step
The Conjugate Pairs Theorem states that when a polynomial has real coefficients, then any complex zeros occur in conjugate pairs. This means that, when $(p +i \ q)$ is a zero of a polynomial function with real number coefficients, then its conjugate $(p –i q)$, is also a zero of the function.
We notice that the function has a degree of $3$. Thus, it has $3$ complex/real zeros. We are given $1$ complex zero and $1$ real zero. Thus, we know that the last zero must be complex, since complex zeros come in pairs. Thus, the remaining zero is: $4-i$, which is the conjugate of $4-i$, by the Conjugate Pairs Theorem.
We can use the zeros to form factors, which we multiply to find the polynomial:
$f(x)=(x-6) [x-(4+i)][x-(4-i)] \\=(x-6) [(x-4)^2 -i^2] \\=(x-6) (x^2-8x+16+1) \\ = x^3-14x^2+65 x -102$
