Answer
yes, $x=-4$ is a solution
Work Step by Step
Using the Remainder Theorem, the remainder when $P(x)=
5x^3+22x^2+x-28
,$ is divided by $(
x+4
)$ is given by $
P(-4)
$.
Substituting $x=-4$ in the function above results to
\begin{align*}
P(-4)&=
5(-4)^3+22(-4)^2+(-4)-28
\\&=
5(-64)+22(16)-4-28
\\&=
-320+352-4-28
\\&=
0
.\end{align*}
Since the remainder is equal to $0$, then $(
x+4
)$ evenly divides $P(x)$. Hence, $
x=-4
$ is a solution of $
5x^3+22x^2+x-28=0
$.
