Roots of: $2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$ This is maybe a stupid question, but I want to find the roots of:

$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$

What that I did:
$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$
So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$ 
My questions:
$1)$ Is there an easy way to see that $x=-8$ is a root too?
$2)$ The degree of this polynomial is $4$, so I should have $4$ roots, and here I have only $3$

Well 
$2(x+2)(x-1)^{3}-3(x-1)^{2}(x+2)^{2}=0 \Rightarrow (x-1)^{2}(x+2)[2(x-1)-3(x+2)]=0 \Rightarrow (x-1)^{2}(x+2)(-x-8)=0 $
From here it should be clear why $x=-8$ is also a root.
This is called factoring.
