Answer
yes, $x=-2$ is a solution
Work Step by Step
Using the Remainder Theorem, the remainder when $P(x)=
x^3-2x^2-3x+10
,$ is divided by $(
x+2
)$ is given by $
P(-2)
$.
Substituting $x=-2$ in the function above results to
\begin{align*}
P(-2)&=
(-2)^3-2(-2)^2-3(-2)+10
\\&=
-8-2(4)-3(-2)+10
\\&=
-8-8+6+10
\\&=
0
.\end{align*}
Since the remainder is equal to $0$, then $(
x+2
)$ evenly divides $P(x)$. Hence, $
x=-2
$ is a solution of $
x^3-2x^2-3x+10=0
$.
