Math help. Please check over my work

Question

UkuleleGirl · Answer

Question: $x^3+2x^2-x-2=0$ 
I wanna know If I did this right. SO i'ma go through this step by step what I did. 
$x^2(x+2)-1(x+2)=0$
$(x^2-1)(x+2)$

I Took the square root of both the -1 and x.$$\sqrt{x^2}=x$$ $$\sqrt{-1}=1$$
$(x+1)(x-1)(x+2)$ 
I set them All equal to 0 and solved.  The end result is: $x=-1,1,-2$
 Feel free to check over my work and please let me know if anything is wrong.

surjithayer · Answer

$$a^2-b^2=(a+b)(a-b)$$
$$x^2-1=x^2-1^2=(x+1)(x-1)$$

surjithayer · Answer

$$\sqrt{-1}=\sqrt{\iota^2}=\pm \iota$$

UkuleleGirl · Answer

@surjithayer wrote:

$$\sqrt{-1}=\sqrt{\iota^2}=\pm \iota$$

uh what

UkuleleGirl · Answer

i just need to know if its right-

surjithayer · Answer

$$\sqrt{-1}\neq 1$$
it is plus minus iota. (an imaginary number)

UkuleleGirl · Answer

I put that.

UkuleleGirl · Answer

just not the sign

surjithayer · Answer

$$just~ write~ x^2-1=x^2-1^2=(x+1)(x-1)$$

UkuleleGirl · Answer

uh...your supposed to group. it says for you to.

UkuleleGirl · Answer

i'm just confused...on what your doing

surjithayer · Answer

you are doing good , nothing to be confused.

UkuleleGirl · Answer

oh yay

avacaplot · Answer

its 1

UkuleleGirl · Answer

Quit lying.

avacaplot · Answer

i swaer