How to perform polynomial long division on 1/(1 - x)? How do I perform polynomial long division on $\frac{1}{1 - x}$ to obtain the sequence $1 + x + x^2 + x^3 + \cdots$?
In this video, the teacher went about it in the following way...
$$
\require{enclose}
\begin{array}{r}
 1 + x + x^2 + x^3 + \cdots \\ 
 1 - x \enclose{longdiv}{\hspace{10pt}1\hspace{85.5pt}} \\
 \underline{-\left(1 - x\right)}\hspace{64.5pt} \\
 x\hspace{68.5pt} \\
 \underline{-\left(x - x^2\right)}\hspace{42pt} \\
 x^2\hspace{46.5pt} \\
 \vdots\hspace{52.5pt}
\end{array}
$$
I have always performed polynomial long division wrt the term of highest degree, e.g., to find $\frac{-7 + 5x + x^2}{-1 + 2x}$, I would do the following...
$$
\require{enclose}
\begin{array}{r}
 x^2 + 5x \hspace{4pt}- 7\hspace{33pt} \\ 
 2x - 1 \enclose{longdiv}{\hspace{10pt}2x^3 + 9x^2 - 19x + 7\hspace{4pt}} \\
 \underline{-\left(2x^3 - x^2\right)}\hspace{50.5pt} \\
 10x^2 - 19x + 7\hspace{4pt} \\
 \underline{-\left(10x^2 - 5x\right)}\hspace{21.5pt} \\
 -14x + 7\hspace{4pt} \\
 \underline{-\left(-14x + 7\right)}\hspace{0pt} \\
 0\hspace{4pt} \\
\end{array}
$$
When should I use the teachers variation of the conventional method?
 A: You do it when $x$ is small compared to $1$.  That is not as silly a remark as it sounds like it is.  If you stop the video's approach part way through you get a remainder term just like you might with the top down approach.  You could write $$\frac 1{1-x}=1+x+x^2+\frac {x^3}{1-x}$$
This is an algebraic fact, valid for all values of $x$ except $x=1$.  When you keep going you get an infinite sum on the right, which is useful as long as it converges.  It converges when $|x| \lt 1$.  If $|x| \ll 1$ the sum converges quickly and you can decide when the error committed by truncating it is acceptable.  
In some problems you know that $x$ is small, in which case this is quite useful.  
If the division is going to come out even, you can do it either way around.  Taking your example
$$\require{enclose}
\begin{array}{r}
 -7 + 5x \hspace{4pt}+x^2\hspace{33pt} \\ 
 -1+2x \enclose{longdiv}{\hspace{10pt}7-19x+9x^2+2x^3\hspace{4pt}} \\
 \underline{7-14x}\hspace{30.5pt} \\
 \hspace {30 pt}- 5x+9x^2\hspace{4pt} \\
 \underline{- 5x+10x^2}\hspace{21.5pt} \\
 -x^2 + 2x^3\hspace{4pt} \\
 \underline{-x^2+2x^3}\hspace{0pt} \\
 0\hspace{4pt} \\
\end{array}$$
where I can't get the spacing as nice as you did, but it gives the same result.
A: Another way to look at it. Start with the known formulas:
$$\begin{align}1-x^2&=(1-x)(1+x)\\
1-x^3&=(1-x)(1+x+x^2)\\
1-x^4&=(1-x)(1+x+x^2+x^3)\\
&\ \ \vdots\\
1-x^n&=(1-x)(1+x+x^2+x^3+\cdots+x^{n-1}) \Rightarrow \\
\frac{1-x^n}{1-x}&=1+x+x^2+x^3+\cdots +x^{n-1}\end{align}$$
If $|x|<1$ and $n\to \infty$, then $x^n\to 0$ and you get the relation:
$$\frac1{1-x}=1+x+x^3+x^4+\cdots$$
