Polynomial division and geometric series By long division $$\dfrac{5x^2+2x-4}{x-2} =5x+12+\color{red}{\dfrac{20}{x-2}}$$
If we continue dividing we get
$$\dfrac{5x^2+2x-4}{x-2} =5x+12+\color{red}{ 20x^{-1}+40x^{-2}+ \dfrac{80x^{-2}}{x-2}}$$
I'm trying to understand two items here 
1) What does the expansion with negative exponents represent  ?
2) If we expand the same using geometric series formula I get a different result (with out negative exponents) :
$$\color{Red}{\dfrac{20}{x-2} = -10\dfrac{1}{1-(x/2)}=-10\left(1+(x/2) + (x/2)^2 + \cdots\right)}  $$
Why did the expansion using long division differ from geometric series ?  

3) Since geometric series only makes sense when $|x/2|\lt 1$, I'm not so sure if the above expansion is valid..

This expansion is the Laurent series of the function $\dfrac{5x^2+2x-4}{x-2}$ around $x = 0$ on the annulus $|x| < 2$.
Actually, you are getting the same result both ways:
\begin{align}
\frac{20}x + \frac{40}{x^2} + \frac{80}{x^2(x-2)} &= \frac{20}x + \frac{40}{x^2} - \frac{40}{x^2}\frac1{1-\frac{x}2}\\
&= \frac{20}x + \frac{40}{x^2} - \frac{40}{x^2}\sum_{n=0}^\infty \frac{x^n}{2^n}\\
&= \frac{20}x + \frac{40}{x^2} - \frac{40}{x^2} - \frac{20}x - 10\sum_{n=2}^\infty \frac{x^{n-2}}{2^{n-2}}\\
&= - 10\sum_{n=0}^\infty \frac{x^{n}}{2^{n}}\\
\end{align}
The Laurent series on $|x| > 2$ is
\begin{align}
\dfrac{5x^2+2x-4}{x-2} &=5x+12+\color{red}{\dfrac{20}{x-2}}\\
&= 5x+12+\color{red}{\frac{20}{x}\frac{1}{1-\frac2x}}\\
&= 5x+12+\color{red}{\frac{20}{x}\sum_{n=0}^\infty \frac{2^n}{x^n}}\\
&= 5x+12+\color{red}{10\sum_{n=1}^\infty \frac{2^n}{x^n}}\\
\end{align}
