Answer
Remainder = $0$,
Solution set =$\{-1,2,3\}$
Work Step by Step
Divide $x^{3}-4x^{2}+x+6$
with $x+1,\qquad (c=-1)$
$\begin{array}{rrrrrrrr}
\underline{-1}| &1 & -4 & 1 & 6 & & & & \\
& & -1 & 5 &-6 & & & & \\
& --& -- &-- &-- & & & & \\
& 1 & -5 & 6 & \fbox{0} & & & & \end{array}$
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$, and the remainder is 0, so
$\displaystyle \frac{x^{3}-4x^{2}+x+6}{(x+1)}=x^{2}-5x+6 $
... the trinomial on the RHS can be factored
(two factors of 6 with sum -5 are $-2$ and $-3$.
Thus, $ x^{2}-5x+6 =(x-2)(x-3)\quad$ and
$x^{3}-4x^{2}+x+6=(x+1)(x-2)(x-3)$
The equation $x^{3}-4x^{2}+x+6=0$ becomes
$(x+1)(x-2)(x-3) =0\quad$ ... apply the zero product principle
Solution set =$\{-1,2,3\}$
