Answer
no, $x=-2$ is NOT a solution
Work Step by Step
Using the Remainder Theorem, the remainder when $P(x)=
x^4+2x^3-3x^2+8x-8
,$ 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(-2)^3-3(-2)^2+8(-2)-8
\\&=
16+2(-8)-3(4)+8(-2)-8
\\&=
16-16-12-16-8
\\&=
-36
.\end{align*}
Since the remainder (equal to $
-36
)$ is not zero, then $(
x+2
)$ DOES NOT evenly divide $P(x)$. Hence, $
x=-2
$ is NOT a solution of $
x^4+2x^3-3x^2+8x=8
$.
