Answer
$4$ and $-2i$
Work Step by Step
RECALL:
If $a+bi$ is a zero of a polynomial function with real number coefficients, then its conjugate $a-bi$ is also a zero of the function.
Thus, one of the missing zeros of the given function is the conjugate of $2i$, which is ${\bf -2i}$.
Thus, two of the three zeros of the polynomial function with real coefficients are:
$2i$ and $-2i$
The third zero is a real number and can be found by factoring the polynomial:
$P(x) = x^3-4x^2+4x-16
\\P(x) = (x^3-4x^2)+(4x-16)
\\P(x) = x^2(x-4)+4(x-4)
\\P(x) = (x-4)(x^2+4)$
Equating each factor to zero gives:
\begin{array}{ccc}
&x-4=0 &\text{or} &x^2+4=0
\\&x=4 &\text{or} &x^2=-4
\end{array}
Thus, the real zero of the function is $4$.
Therefore, the missing zeros are:
$4$ and $-2i$
