Answer
$-i, i, -5i,$ and $5i.$
Work Step by Step
First, because all coefficients of f(x) are real numbers, the Conjugate Zeros Theorem applies.
If $i$ is a zero, so is its conjugate, $-i.$
If k is a zero of f, then, by the factor theorem, (x-k) is a factor of f(x).
Synthetic division gives us a quotient q(x), with degree one less than f(x).
and we know the remainder is zero,
so $f(x)=(x-k)\cdot q(x).$
We use this now
$k=i$,, note the missing $x^{3}$ and $x$ terms
$\left\{\begin{array}{lllllll}
i & | & 1 & 0 & 26 & 0 & 25\\
& & & i & -1 & 25i & -25\\
& & -- & -- & -- & -- & --\\
& & 1 & i & 25 & 25i & \fbox{$0$}
\end{array}\right.$
$f(x)=(x-i)(x^{3}+ix^{2}+25x+25i)$
Now, we use the fact that $-i$ is also a zero.
Apply synthetic division on the quotient.
$\left\{\begin{array}{llllll}
-i & | & 1 & i & 25 & 25i\\
& & & -i & 0 & -25i\\
& & -- & -- & -- & --\\
& & 1 & 0 & 25 & \fbox{$0$}
\end{array}\right.$
$f(x)=(x+i)(x-i)(x^{2}+25)$
The zeros of $x^{2}+25:$
$x^{2}+25=0$
$x^{2}=-25$
$x=\pm 5i$
The other zeros (other than the given) are
$-i,\quad -5i$ and $5i.$