Answer
NOT a factor
Work Step by Step
Equat$x-3,$ to zero then solve for $x,$:
\begin{align*}
x-3&=0
\\
x&=3
\end{align*}
Substituting $x=
3
$ in the given expression, $
x^3+4x^2+x-6
,$ results to
\begin{align*}
&
(3)^3+4(3)^2+(3)-6
\\&=
27+4(9)+3-6
\\&=
27+36+3-6
\\&=
60
.\end{align*}
Since the substitution above is NOT equal to zero, then the remainder is NOT zero. By the Factor Theorem, $
x-3
$ is not a factor of $x^3+4x^2+x-6$.