Answer
The remaining zeros of the function are:
$\color{blue}{\bf 2i, \frac{1}{2}, -3}$
Work Step by Step
The polynomial function's degree is four so it has four zeros.
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 ${\bf 2i}$.
Two of the four zeros of the polynomial function with real coefficients are:
$-2i$ and $2i$
This means that $(x-2i)$ and $(x+2i)$ are both factors of $P(x)$.
Multiply these two to obtain:
$(x-2i)(x+2i) = x^2-4i^2 = x^2-4(-1) = x^2+4$
Divide $P(x)$ by $x^2+4$ using long division to obtain $2x^2+5x-3$.
(refer to the attached image below for the actual long division)
Factor $2x^2+5x-3$ to obtain:
$2x^2+5x-3=(2x-1)(x+3)$
Thus, the completely factored form of $P(x)$ is:
$P(x) = (x^2+4)(2x-1)(x+3)$
Equate each factor to zero, then solve each equation to obtain:
\begin{array}{ccccc}
&x^2+4=0 &\text{or} &2x-1=0 &\text{or} &x+3=0
\\&x^2=-4 &\text{or} &2x=1 &\text{or} &x=-3
\\&x^2=-4 &\text{or} &x=\frac{1}{2} &\text{or} &x=-3
\end{array}
The first factor's solutions are the two complex zeros of the function.
Thus, the real zeros are $x=\left\{\frac{1}{2}, -3\right\}$.
Therefore, the other zero of the given function are:
$\color{blue}{\bf 2i, \frac{1}{2}, -3}$
