From the rational root test, the potential rational roots are:
By trial and error we find that is a root: , i.e. is one of the factors.
Long divide to get the blue factor:
Now factor again and you should be on your way.
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