Answer
$-2i, 4 $
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.
Its degree is $3$, hence it has $3$ complex zeros. According to the Conjugate Pair Theorem since $2i$ is a complex zero, $-2i$ is also a complex zero.
Assume $r$ is the last zero. Then we know that $(x-2i)(x-(-2i))(x-r)=x^3-4x^2+4x-16\\(x-2i)(x+2i)(x-r)=x^3-4x^2+4x-16\\(x^2+4)(x-r)=x^3-4x^2+4x-16\\x^3-rx^2+4x-4r=x^3-4x^2+4x-16\\-rx^2+4x-4r=-4x^2+4x-16\\-rx^2-4r=-4x^2-16\\-r(x^2-4)=-4(x^2-4)\\-r=-4\\r=4$
