Answer
Not a factor.
Work Step by Step
Using synthetic division =, we divide $f(x)$ by $x-k$ to obtain
the quotient $q(x)$ and remainder $r(x)$ such that
$f(x)=(x-k)\cdot q(x)+r(x)$
$k=-1$
$\left\{\begin{array}{llllllll}
-1 & | & 2 & 5 & -8 & 3 & 13 & \\
& & & -2 & -3 & 11 & -14 & \\
& & -- & -- & -- & -- & -- & \\
& & 2 & 3 & -11 & 14 & \fbox{$-1$} &
\end{array}\right.$
The last row gives us the coefficients of the quotient
and the remainder (boxed).
By the remainder theorem, $f(-1)=-1$, so
$k=-1$ is not a zero of f.
Apply the factor theorem, which states that the following statements :
$\left\{\begin{array}{l}
\text{$k$ is a zero of f}\\
\text{$(x-k)$ is a factor of f}
\end{array}\right.\quad $ are equivalent.
Since $-1$ is not a zero of f,
$(x-k)$ = $(x+1)$ is not a factor of f.