Answer
$4$ and $-2i$
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 a real number of the coefficients, then its conjugate $(p –i q)$, is also a zero of the function.
Since we are told that $2i$ is one zero, the other zero must be $-2i$.
The third zero must be a real number that can be computed by factoring the polynomial: $f(x) = x^3-4x^2+4x-16 \\ = (x^3-4x^2)+(4x-16) \\= (x-4)(x^2+4)$
By the zero product property, we have: $x=4 $ and $x^2=-4$. Thus, the real zero of the function is $4$. This implies that the missing zeros are: $4$ and $-2i$
