Subtracting rational functions I'm trying to find out how to solve this:
$$\frac{x-2}{x^2+2x} - \frac{x+2}{x^2-2x} - \frac{4x}{x^2-4}$$
The answer is $\displaystyle \frac{-4}{x-2}$
What is this called? And is there any video of it on http://www.khanacademy.org ?
Thanks!

Added: The method is to transform the sum of the given rational fractions (the numerator and denominator consists of polynomials) into a
single equivalent fraction.
The properties used are


*

*$$\frac{A(x)}{B(x)}=\frac{A(x)P(x)}{B(x)P(x)}\qquad\text{for  }P(x)\neq 0.$$

*$$\frac{A(x)}{B(x)}\pm \frac{C(x)}{D(x)}=\frac{A(x)D(x)\pm B(x)C(x)}{B(x)D(x)}.$$

I would calculate as follows, starting with the factorization of denominators
$$\begin{eqnarray*}
&&\frac{x-2}{x^{2}+2x}-\frac{x+2}{x^{2}-2x}-\frac{4x}{x^{2}-4} \\
&=&\frac{x-2}{x(x+2)}-\frac{x+2}{x\left( x-2\right) }-\frac{4x}{\left(
x-2\right) \left( x+2\right) } \\
&=&\frac{\left( x-2\right) ^{2}}{x(x+2)\left( x-2\right) }-\frac{\left(
x+2\right) ^{2}}{x\left( x-2\right) \left( x+2\right) }-\frac{4x^{2}}{%
x\left( x-2\right) \left( x+2\right) } \\
&=&\frac{\left( x-2\right) ^{2}-\left( x+2\right) ^{2}-4x^{2}}{x\left(
x-2\right) \left( x+2\right) } \\
&=&\frac{x^{2}-4x+4-x^{2}-4x-4-4x^{2}}{x\left( x-2\right) \left( x+2\right) }
\\
&=&\frac{-4x^{2}-8x}{x\left( x-2\right) \left( x+2\right) } \\
&=&-\frac{4x\left( x+2\right) }{x\left( x-2\right) \left( x+2\right) } \\
&=&-\frac{4}{x-2}.
\end{eqnarray*}$$
so that I reduce  the rational fractions to a common denominator first to allow me to add (subtract) them, and simplify the numerator thereafter. Finally I divide by the common factors to both the numerator and denominator. This is valid iff $x\ne -2$ and $x\ne 0$, because you cannot divide by zero. Also for $x=2$ the fraction is not defined.
See "Simplifying Rational Expressions 1, 2 and 3", "Adding and Subtracting Rational Expressions 1, 2 and 3"  on http://www.khanacademy.org.
