Question about inverse in power formal series

Let the ring $R[X]$ of all power formal series $\sum_{i=-\infty}^{\infty}{a_iX^i}$, (where $a_i$ are coefficients in finite field $K$) in which $a_i=0$, for every $i>0$ with at most a finite number. Observe that $K[X]$, the ring of polynomials in $X$, is embedded in $R[X]$ in natural manner.

My question is: Let $g(X)$ and $f(X) \in R[X]$, with degree $n$ and $n-1$ respect. How i get the g(X) inverse with indeterminants $X^{-i}$?. Because in my text say

$f(X) = ((a_0X^{-1}+a_1X^{-2}+...+a_{n-1}X^{-n}) + b_nX^{-(n+1)}+...)g(X)$.

pdta: I think that $f(X)=f(X)g(X)^{-1}g(X)$ then $f(X)g(X)^{-1} = ((a_0X^{-1}+a_1X^{-2}+...+a_{n-1}X^{-n}) + b_nX^{-(n+1)}+...)$, but my problem is the inverse of $g(X)$

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The hint is: long division. In these days of calculators, they still teach that, don't they? – GEdgar Feb 27 '12 at 21:28
no understand GEdgar – juaninf Feb 27 '12 at 23:48
If you want to get someone's attention, you have to write @GEdgar – Gerry Myerson Feb 28 '12 at 11:41

Divide $x+1$ by $x^2-x-1$, express the answer as a Laurent series at $x=\infty$.