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

Mar 15, 2016

$\frac{{x}^{3} + {x}^{2} - 2 x}{{x}^{3} + 2 {x}^{2} - x - 2} = 1 - \frac{1}{x + 1}$

with exclusions $x \ne 1$ and $x \ne - 2$

#### Explanation:

$\frac{{x}^{3} + {x}^{2} - 2 x}{{x}^{3} + 2 {x}^{2} - x - 2}$

$= \frac{x \left({x}^{2} + x - 2\right)}{{x}^{2} \left(x + 2\right) - 1 \left(x + 2\right)}$

$= \frac{x \left(x - 1\right) \left(x + 2\right)}{\left({x}^{2} - 1\right) \left(x + 2\right)}$

$= \frac{x \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 2\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \left(x + 1\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 2\right)}}}}$

$= \frac{x}{x + 1}$

$= \frac{x + 1 - 1}{x + 1}$

$= 1 - \frac{1}{x + 1}$

with exclusions $x \ne 1$ and $x \ne - 2$

$\textcolor{w h i t e}{}$
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.

Mar 16, 2016

By factoring and factoring by grouping
$\frac{x}{x + 1}$

#### Explanation:

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
$\frac{x \left({x}^{2} + x - 2\right)}{\left({x}^{3} + 2 {x}^{2}\right) - 1 \left(x + 2\right)}$
{x(x+2)(x-1)}/(x^2(color(blue)(x+2))-1(color(blue)(x+2))
$\frac{x \left(x + 2\right) \left(x - 1\right)}{\textcolor{b l u e}{\left(x + 2\right)} \left({x}^{2} - 1\right)}$
{x(cancel(x+2))cancel((x-1))}/{color(blue)((cancel(x+2))(x+1)(cancel(x-1))}

3. Final Asnwer:
$\frac{x}{x + 1}$