Answer
The zeros are
$2-i$ , (which is given,)
$2+i$ and
$3.$
Work Step by Step
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 first.
$k=2-i$, ,
$\left\{\begin{array}{llllll}
2-i & | & 1 & -7 & 17 & -15\\
& & & 2-i & -11+3i & 15\\
& & -- & -- & -- & --\\
& & 1 & -5-i & 6+3i & \fbox{$0$}
\end{array}\right. \qquad\left(\begin{array}{lll}
(2-i)(-5-i) & | & (2-i)(6+3i)\\
=-10+3i+i^{2} & | & =12-3i^{2}\\
=-11+3i & | & =15
\end{array}\right)$
$f(x)=[x-(2-i)][x^{2}+(-5-i)x+(6+3i)]$
We now use the Conjugate Zeros Theorem, by which,
since $z=2-i $ is a zero,
its conjugate, $2+i$ is also a zero.
So, we apply synthetic division on the quotient.
$\left\{\begin{array}{llllll}
2+i & | & 1 & -5-i & 6+3i & \\
& & & 2+i & -6-3i & \\
& & -- & -- & -- & \\
& & 1 & -3 & \fbox{$0$} &
\end{array}\right.$
$f(x) =[x-(2-i)]\cdot[x-(2+i)]\cdot(x-3)$
By the factor theorem,
$x-3$ is a factor of $f(x) \Rightarrow 3$ is a zero.
The zeros are
$2-i$ , (which is given,)
$2+i$ and
$3.$
