#### Answer

$\color{blue}{\bf -3i, \frac{1}{3}, -2}$

#### 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 $3i$, which is ${\bf -3i}$.
Two of the four zeros of the polynomial function with real coefficients are:
$-3i$ and $3i$
This means that $(x-3i)$ and $(x+3i)$ are both factors of $P(x)$.
Multiply these two to obtain:
$(x-3i)(x+3i) = x^2-9i^2 = x^2-9(-1) = x^2+9$
Divide $P(x)$ by $x^2+9$ using long division to obtain $3x^2+5x-2$.
(refer to the attached image below for the actual long division)
Factor $3x^2+5x-2$ to obtain:
$3x^2+5x-2=(3x-1)(x+2)$
Thus, the completely factored form of $P(x)$ is:
$P(x) = (x^2+9)(3x-1)(x+2)$
Equate each factor to zero, then solve each equation to obtain:
\begin{array}{ccccc}
&x^2+9=0 &\text{or} &3x-1=0 &\text{or} &x+2=0
\\&x^2=-9 &\text{or} &3x=1 &\text{or} &x=-2
\\&x^2=-9 &\text{or} &x=\frac{1}{3} &\text{or} &x=-2
\end{array}
The first factor's solutions are the two complex zeros of the function.
Thus, the real zeros are $x=\left\{\frac{1}{3}, -2\right\}$.
Therefore, the other zero of the given function are:
$\color{blue}{\bf -3i, \frac{1}{3}, -2}$