Answer
The remaining zeros are:
$-3$ and $5i$
Work Step by Step
The polynomial function's degree is three so it has three 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 $-5i$, which is ${\bf 5i}$.
Thus, two of the three zeros of the polynomial function with real coefficients are:
$-5i$ and $5i$
The third zero is a real number and can be found by factoring the polynomial:
$P(x) = x^3+3x^2+25x+75
\\P(x) = (x^3+3x^2)+(25x+75)
\\P(x) = x^2(x+3)+25(x+3)
\\P(x) = (x+3)(x^2+25)$
Equating each factor to zero gives:
\begin{array}{ccc}
&x+3=0 &\text{or} &x^2+25=0
\\&x=-3 &\text{or} &x^2=-25
\end{array}
Thus, the real zero of the function is $-3$.
Therefore, the missing zeros are:
$-3$ and $5i$
