Answer
Not a factor.
Work Step by Step
Using synthetic division =, we divide $f(x)$ with $x-k$ to obtain
the quotient $q(x)$ and remainder $r(x)$ such that
$f(x)=(x-k)\cdot q(x)+r(x)$
$k=1$
$\left\{\begin{array}{llllllll}
1 & | & -3 & 1 & -5 & 2 & 4 & \\
& & & -3 & -2 & -7 & -5 & \\
& & -- & -- & -- & -- & -- & \\
& & -3 & -2 & -7 & -5 & \fbox{$-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}\\
\text{(x-k) is a factor of f}
\end{array}\right.\quad $ are equivalent.
Since $1$ is not a zero,
$(x-k)$ = $(x-1)$ is not a factor of the polynomial.