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)#