Why this way of dision is not correct; that is, changing the order of the terms of divisor as and when required? Even though $1+x$ is same as $x+1$
Check:
(quotient)X(divisor)+(remainder) gives$[(1-1+1-1)(1+x)]+x=x$
Same thing happens when we divide x by (1-x) the above way.
My thought:
May be it is not helping because in one step divisor is removed from the dividend and in next step it is added to the dividend, so both steps nullify each other and we are left the original dividend!, but then there should be a rule that you can not change the order of the terms of the divisor while performing algebraic division!
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$\begingroup$There is a rule about changing the order of terms, but it's not quite what you think.
You have to decide whether the indeterminate variable $x$ is "big" or if it is "small". that is, whether $x$ should be considered a larger or smaller polynomial than $1$.
If $x$ is "big", then the size of a polynomial is its degree. If $x$ is "small", then if $x^m$ is the smallest power of $x$ appearing in it, you consider the size of the polynomial to be $-m$.
The order you write down the terms of your polynomial aren't really relevant: though if you stick with a consistent convention, the terms will be sorted based on their size. This might have prevented you from figuring out what's 'really' going on.
Each step of long division should involve the largest terms of the two polynomials. If you consider $x$ to be "big", then your second step was in error: when dividing $-1$ by $1+x$, the largest terms are $-1$ and $x$ respectively, so you should have gotten $-x^{-1}$ as the quotient and $x^{-1}$ as the remainder. Repeating will derive the series
$$ \frac{x}{1+x} = 1 - \frac{1}{x} + \frac{1}{x^2} - \cdots $$
However, if you consider $x$ to be "small", then your first step was in error: when dividing $x$ by $1+x$, the largest terms are $x$ and $1$ respectively, so your quotient should have been $x$, and your remainder $-x^2$. Repeating will derive the series
$$ \frac{x}{1+x} = x - x^2 + x^3 - x^4 + \cdots $$
Incidentally, there are other choices than these for size. If $a$ is any scalar, then you can treat $(x-a)$ as smaller than $1$. i.e. the size of a polynomial is computed by rewriting your polynomial $(x-a)$ rather than in terms of $x$. Then if $(x-a)^m$ is the smallest power of $(x-a)$ appearing, you consider the size of the polynomial to be $-m$.
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