Answer
yes, $x=−2$ is a solution
Work Step by Step
Using the Remainder Theorem, the remainder when $P(x)=
x^4-x^3-6x^2+5x+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)^4-(−2)^3-6(−2)^2+5(−2)+10
\\&=
16-(−8)-6(4)+5(−2)+10
\\&=
16+8-24-10+10
\\&=
0
.\end{align*}
Since the remainder is $0$, then $(
x+2
$) evenly divides $P(x)$. Hence, $
x=−2
$ is a solution of $
x^4-x^3-6x^2+5x=-10
$.
