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 $3$
 A: Use the Distributive Property.
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2$$
$$=\underbrace{(x+2)(x-1)^2}_{\text{common factor}}\left(2(x-1)\right)-\underbrace{(x+2)(x-1)^2}_{\text{common factor}}(3(x+2))$$
$$=(x+2)(x-1)^2\left(2(x-1)-3(x+2)\right)$$
$$=(x+2)(x-1)^2(-(x+8))$$
A: First advice: when you are looking for roots of a polynomial, factorize as much as you can. You have $(x-1)^2$ and $(x-1)^3$, so you  can factorize  $(x-1)^2$.   You have $(x+2)$ and $(x+2)^2$, so you  can factorize  $(x+2)$. Thus you get:
$$(x+2)(x-1)^2 (2(x-1)-3(x+2))\,. $$ So:


*

*For question 1,  the  $-8$ root should be evident (develop the  last term)

*For question 2: look at powers, $(x-1)^2$ means that $1$ is a root, twice. So your roots are $-8$, $-2$ and  $1$ and $1$ again.

A: Ніnt:
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=- \left( x+8 \right)  \left( x+2 \right)  \left( x-1 \right) ^{2}.$$
A: $$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$
$$(x-1)^2(x+2)\left[2(x-1)-3(x+2)\right]=0$$
$$(x-1)^2(x+2)(-x-8)=0$$
$$(x-1)^2(x+2)(x+8)=0$$
So this polynomial of degree $4$ has $4$ real roots i.e. $1,1,-2,-8$.
A: 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.
