This generalization is useless for application but it will lead to a formula for polynomial division. If we have $f/g$ which is $$M(x)=\frac{\sum_{n=0}^{l=m+v}{{a_n}x^{n}}}{\sum_{n=0}^{l=m}{b_n}x^{n}}=\frac{a_0+a_1x+a_2{x^2}...+a_{m+v}{x^{m+v}}}{b_0+b_1x+b_2x^2...+b_mx^m}$$ Just take the taylor series at infinity. Substitute $1/x$ in $x$ then multiply the numerator and denominator by $\frac{1}{x^{m+v}}$ and factor the denominator by $x^v$ to get. $$\left(\frac{1}{x^v}\right)\frac{a_0+a_1x+a_2{x^2}...+a_{m+v}{x^{m+v}}}{b_0+b_1x+b_2x^2...+b_mx^m}=\left(\frac{1}{x^v}\right)L(x)$$

Where $$L(x)=\frac{a_{m+v}+a_{m+v-1}x+a_{m+v-2} {x^2}....+a_{0}{x^{m+v}}}{b_{m}+b_{m-1}x+b_{2}x^2...+b_0x^{m}}$$

Where v is the difference between the highest exponent degrees of a and b.

Then if we take the taylor series of $L(x)$, so the series converges to infinity and substiute $\frac{1}{x}$ for all the $x$'s. $${x^v}\left(L(0)+{1!}L^{1}(0){\frac{1}{x}}+\frac{1}{2!}L^{2}(0){\frac{1}{x^2}}..+\frac{1}{v!}L^{v}(0){\frac{1}{x^v}}\right)$$ $$L(0){x^v}+{1!}L^{1}(0){x^{v-1}}+\frac{1}{2!}L^{2}(0){x^{v-2}}..+\frac{1}{v!}L^{v}(0)$$ So this can be generalized as $$q(x)=\left(\sum_{i={-v}}^{0}\frac{L^{i+v}(0){x^{-i}}}{\left(i+v\right)!}\right)$$

The result this quotoient we can get...

$$\left(\frac{a_{m+v}}{b_m}\right)x^v+\left(\frac{a_{m+v-1}b_{m-1}}{b_m}+\frac{a_{m+v}b_{m-1}}{{b_m}^2}\right)x^{v-1}+\frac{1}{2!}\left(\frac{2{a_{m+v-2}}}{b_m}+\frac{{2}{a_{m+v}}{b_{m-1}}^{2}}{{b_{m}}^{3}}-\frac{{2}{a_{m+v}}{b_{m-2}}}{{b_{m}}^{2}}+\frac{{2}{a_{m+v-1}}{b_{m-1}}}{{b_{m}}^{2}}\right)x^{v-2}+\frac{1}{3!}...$$

Now all you have to do is take the division formula for the remainder $$\frac{a(x)}{b(x)}={q(x)}+\frac{r(x)}{b(x)}$$ $${a(x)}={q(x)}{b(x)}+{r(x)}$$ $${a(x)}-{q(x)}{b(x)}={r(x)}$$ But I can't expand the remainder because too tedious and complicated.

So the complete form is.. $$M(x)=\left(\sum_{i={-v}}^{0}\frac{L^{i+v}(0){x^{-i}}}{\left(i+v\right)!}\right)+\frac{\sum_{n=0}^{l=m+v}{{a_n}x^{n}}-\sum_{i={-v}}^{0}\frac{L^{i+v}(0){x^{-i}}}{\left(i+v\right)!}{\sum_{n=0}^{l=m}{b_n}x^{n}}}{\sum_{n=0}^{l=m}{b_n}x^{n}}$$ It isn't pretty but it is a formula.

Not sure if this will suffice. I was curious myself to find a formula specifically for the division of $n$th degree polynomials by a linear factor. This is the formula, where $n$ is the degree of $P$. Clearly not to supple a result. $$P(x)/(x+c)=\sum^n_{i=1}(\sum^i_{j=1}(-c)^{i-j} a_{n-j+1}))x^{n-j}+P(-c)/(x+c)$$

each term $a_{n-j+1}$ refers to a coefficient of $P(x)=a_nx^n+a_{n-1}x^{n-1}+. . .+a_1x+a_0$

Polynomial division is equivalent to deconvolution using an IIR filter with an impulse as input, where 'filter' is meant in the signal processing sense and not the set theory sense. (If there is no remainder, then the IIR filter becomes the FIR filter.)

Let $f = a_0 x^{N-1} + a_i x^{N-2} + \cdots a_{N-1}x^0$ and likewise for $g$.For $\mathbf{b}\in \mathbb{R}^M$ and $\mathbf{a} \in \mathbb{R}^N$ such that $a_0=1$, the polynomial division is given by

$$ y_n = \sum_{k=0}^{M-1}{ \delta_{n-k} b_k } - \sum_{i=1}^{\min(n,N-1)} y_{n-i} a_i $$

where

$$ \delta_k = \begin{cases} 1, & k = 0 \\ 0, & k \neq 0 \end{cases} $$

The first $M-N+1$ terms of ${y_n}$ will be the coefficients of the quotient and the remaining non-zero terms will be the coefficients of the remainder.