Solve this cubic equation for x:x^3-4x^2+x+6=0
From the rational root test, the potential rational roots are: $\displaystyle \pm 1, \ \pm 2, \ \pm 3, \ \pm 6$
By trial and error we find that $\displaystyle x=2$ is a root: $\displaystyle (2)^3-4(2)^2+(2)+6 = 0$, i.e. $\displaystyle (x-2)$ is one of the factors.
Long divide to get the blue factor: $\displaystyle x^3 - 4x^2 + x + 6 = (x-2){\color{blue}(x^2 - 2x - 3)}$
Now factor again and you should be on your way.
$\displaystyle x^3-4x^2+x+6=0$
Using synthetic division we get:
+1 -4 +1 +6
-1 0 -1 +5 -6
+1 -5 +6 0
so, -1 is a root of the given equation,so the given equ becomes:
(x+1)(x^2-5x+6)=0
which implies (x+1)(x^2-2x-3x+6)=0
which implies (x+1)(x(x-2)-3(x-2))=0
which implies (x+1)(x-2)(x-3)=0
which implies x+1=0 ,x-2=0 ,x-3=0
which implies x=-1, x=2,x=3