How do you simplify `(x^3+x^2-2x)/(x^3+2x^2-x-2)`?

`(x^3+x^2-2x)/(x^3+2x^2-x-2)`

`=(x(x^2+x-2))/(x^2(x+2)-1(x+2))`

`=(x(x-1)(x+2))/((x^2-1)(x+2))`

`=(x color(red)(cancel(color(black)((x-1))))color(red)(cancel(color(black)((x+2)))))/(color(red)(cancel(color(black)((x-1))))(x+1)color(red)(cancel(color(black)((x+2)))))`

`=x/(x+1)`

`=(x+1-1)/(x+1)`

`=1-1/(x+1)`

with exclusions `x != 1` and `x != -2`

`color(white)()`
These exclusions are required, because when `x=1` or `x=-2` both the numerator and denominator of the original rational expression are zero, so the quotient is undefined, but the simplified expression is well behaved at these values of `x`.

Note however that we do not need to specify `x=-1` as an excluded value of our simplification, because it is equally a simple pole of both the original and simplified expressions.

1. Factor both your numerator and denominator
   `(x^3+x^2-2x)/(x^3+2x^2color(red)(-x-2)`
   `{x(x^2+x-2)}/{(x^3+2x^2)color(red)(-1(x+2)}`

2. Again, factor the remaining terms
   `{x(x^2+x-2)}/((x^3+2x^2)-1(x+2))`
   `{x(x+2)(x-1)}/(x^2(color(blue)(x+2))-1(color(blue)(x+2))`
   `{x(x+2)(x-1)}/{color(blue)((x+2))(x^2-1)}`
   `{x(cancel(x+2))cancel((x-1))}/{color(blue)((cancel(x+2))(x+1)(cancel(x-1))}`

3. Final Asnwer:
   `(x)/(x+1)`