Answer
$f(x)= (x+1)^{3}(2x-5)$
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.
$k=-1$
$\left\{\begin{array}{lllllll}
-1 & | & 2 & 1 & -9 & -13 & -5\\
& & & -2 & 1 & 8 & 5\\
& & -- & -- & -- & -- & --\\
& & 2 & -1 & -8 & -5 & \fbox{$0$}
\end{array}\right.$
$f(x)= (x+1)(2x^{3}-x^{2}-8x-5)$
$k=-1$ has multiplicity $3$, so $(x+1)$ is a factor of $(2x^{3}-x^{2}-8x-5)$.
We divide once more:
$\left\{\begin{array}{lllllll}
-1 & | & 2 & -1 & -8 & -5 & \\
& & & -2 & 3 & 5 & \\
& & -- & -- & -- & -- & --\\
& & 2 & -3 & -5 & \fbox{$0$} &
\end{array}\right.$
$f(x)= (x+1)(x+1)(x^{2}-2x-5)$
$k=-1$ has multiplicity $3$, so $(x+1)$ is a factor of $(x^{2}-2x-5)$.
We divide once more:
$\left\{\begin{array}{lllllll}
-1 & | & 2 & -3 & -5 & & \\
& & & -2 & 5 & & \\
& & -- & -- & -- & -- & --\\
& & 2 & -5 & \fbox{$0$} & &
\end{array}\right.$
So,
$f(x)= (x+1)^{3}(2x-5)$
