So, I like to factor this sort of problm using synthetic division. First thing's first, let's set up this problem.

#color(white)(-2)|##1color(white)(000)2color(white)(000)-1color(white)(000)-2#

#color(white)(-2)|#

#-2|#*_**_**_**_**_**_**_**_*__

We bring down the #1#, which gives us this:

#color(white)(-2)|##1color(white)(000)2color(white)(000)-1color(white)(000)-2#

#color(white)(-2)|#

#-2|#*_**_**_**_**_**_**_**_*__

#color(white)(0)color(white)(000)1#

We then multiply the #-2# by the #1# that we brought down to give us #-2#. That value is brought up to the next row, which brings us to this:

#color(white)(-2)|##1color(white)(000)2color(white)(000)-1color(white)(000)-2#

#color(white)(-2)|##color(white)(00)-2#

#-2|#*_**_**_**_**_**_**_**_*__

#color(white)(0)color(white)(000)1#

From here, we just add the #2# to the #-2# and end up at this:

#color(white)(-2)|##1color(white)(000)2color(white)(000)-1color(white)(000)-2#

#color(white)(-2)|##color(white)(00)-2#

#-2|#*_**_**_**_**_**_**_**_*__

#color(white)(0)color(white)(000)1color(white)(000)0#

If we continue this system, we end up with this:

#color(white)(-2)|##1color(white)(000)2color(white)(000)-1color(white)(000)-2#

#color(white)(-2)|##color(white)(00)-2color(white)(0000)0color(white)(000000)2#

#-2|#*_**_**_**_**_**_*__

#color(white)(0)color(white)(000)1color(white)(000)0color(white)(00)-1color(white)(000000)0#

So now we have #x=-2# or #x+2=0#, or #(x+2)# and #(x^2-1)#. This can be simplified to just #(x+1)(x-1)#.

We now have the factors of #x^3+2x^2-x-2#, which are #(x+2)(x+1)(x-1)#, which can be rewritten as: #x=-2,-1,1#.

So let's just double check that we found all the zeros of #x^3+2x^2-x-2# be graphing it

graph{x^3+2x^2-x-2}

And look, the three #x#-intercepts are what we said, #x=1,-1,-2#.