There's a "formula" for how to multiply polynomials, but is there one for dividing them? There probably is, anyone could deduce one with enough time. Do you know a formula or could provide one? By formula I mean, given $f=\sum^m a_k x^k,g=\sum^n b_k x_k$, a formula for $q,r$ in $f=gq+r$.
(If there's any confusion, no, this is not a question on how to do polynomial division.)
Edit: Since there seems to be so much confusion about what I mean, here's an example: Let $m=4$, $n=2$, $a_m=b_n=1$. Then $f/g=q+r/g$ where
$q=x^2+(a_3-b_1)x+(a_2-b_0-a_3b_1+b_1^2)$
$r=(a_0-a_2b_0+b_0^2+a_3b_0b_1-b_0b_1^2)+(a_1-a_3b_0-a_2b_1+2b_0b_1+a_3b_1^2-b_1^3)x$
This is a general formula: Dividing a general monic quartic by a general monic quadratic will always have this form.
Best Answer
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.