Answer
no, $x=-2$ is NOT a solution
Work Step by Step
Using the Remainder Theorem, the remainder when $P(x)=
x^3-3x^2-x+10
,$ is divided by $(
x+2
)$ is given by $
P(-2)
$.
Substituting $x=-2$ in the function above results to
\begin{align*}
P(-2)&=
(-2)^3-3(-2)^2-(-2)+10
\\&=
-8-3(4)-(-2)+10
\\&=
-8-12+2+10
\\&=
-8
.\end{align*}
Since the remainder (equal to $
-8
)$ is not zero, then $(
x+2
)$ DOES NOT evenly divide $P(x)$. Hence, $
x=-2
$ is NOT a solution of $
x^3-3x^2-x+10=0
$.
