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It all started while I was trying to calculate the intersection points between two possibly rotated ellipses, each characterised by an equation like the following: $$\frac{((x-h) \cos A + (y-k) \sin A)^2}{a^2}+\frac{((x-h) \sin A - (y-k) \cos A)^2}{b^2}=1$$ where $(h, k)$ reflect the location of its centre, $a$ and $b$ reflect the lengths of the horizontal and vertical axes prior to rotation, and $A$ reflects the rotation angle in radians.

Trying to follow the suggestions found on this SE answer and solving for $t$, at some point I arrive to the following polynomial division:

$$\frac{\alpha t^8+\beta t^7+\gamma t^6+\delta t^5+\epsilon t^4+\zeta t^3+\eta t^2+\theta t+\iota}{\kappa t^8+\lambda t^6+\mu t^4+\nu t^2+\xi}$$ where the greek coefficients stand for constant expressions in terms of $h$, $k$, $a$, $b$ and $A$.

At this point, I am considering plugging in the expressions at the coefficients, and then trying to factor things out in the resulting equation, but that looks messy because the constant expressions are not exactly simple. This division, however, looks much more simple, so I am also wondering whether I might be able to divide one polynomial into the other and get a lower degree polynomial where I could then plug in the constant expressions and carry on solving for t in terms of $h$, $k$, $a$, $b$ and $A$.

Now, it has been a while since I even looked at a polynomial, nevermind simplifying it, so I have been searching around to refresh my memory regarding the division of polynomials. But all the examples I have found so far have been rather cumbersome procedures with numerical coefficients, so my question is, is there a general method for the division of polynomials with unknown coefficients?

Thanks!

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  • $\begingroup$ My first though is no, not in any useful form. Is there a general method for division of unknown integers? $\endgroup$ – basket Sep 18 '16 at 17:15
  • $\begingroup$ Thanks @basket, but if I knew the answer to your question, I wouldn't be asking mine :-( $\endgroup$ – Carvo Loco Sep 18 '16 at 17:43
  • $\begingroup$ I guess @basket was asking a rhethorical question; with no as the implied answer. Perhaps you can avoid the division? What's the equation to go with that fraction? Perhaps you can just multiply that equation by the denominator of the fraction to end up in a “simple” 8th degree polynomial. Still far higher than the 4th degree claimed by the answer you reference, so presumably you did something wrong along the way to that equation. $\endgroup$ – MvG Sep 19 '16 at 7:00
  • $\begingroup$ Hello @MvG, thanks for your suggestions. I was hoping that my expression would be eventually simplified to a 4th degree one, but you are right, I am probably carrying along an early mistake. Nonetheless, I was wondering whether, just like I can calculate that $(Ax^2 + Bx + C)(Dx + E)=ADx^3 + (AE + BD)x^2 + (BE + CD)x + CE$, I might be able to do something similar with $(Ax^2 + Bx + C) / (Dx + E)$. Thanks for your comments, would you or basket like to post an answer that I can mark as accepted? $\endgroup$ – Carvo Loco Sep 19 '16 at 21:33
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In general there is no canonical way to simplify a polynomial division if you don't actually know the coefficients. I can't think of a way where knowing that the off coefficients of the denominator are zero would be of any use.

Perhaps you can avoid the division? What's the equation to go with that fraction? Perhaps you can just multiply that equation by the denominator of the fraction to end up in a “simple” 8th degree polynomial. For example,

$$\frac{\alpha t^8+\beta t^7+\gamma t^6+\delta t^5+\epsilon t^4+\zeta t^3+\eta t^2+\theta t+\iota}{\kappa t^8+\lambda t^6+\mu t^4+\nu t^2+\xi}=\rho$$

simplifies to

$$(\alpha-\rho\kappa) t^8+\beta t^7+ (\gamma-\rho\lambda) t^6+\delta t^5+ (\epsilon-\rho\mu) t^4+\zeta t^3+ (\eta-\rho\nu) t^2+\theta t+ (\iota-\rho\xi)=0$$

Still far higher than the 4th degree claimed by the answer you reference, so presumably you did something wrong along the way to that equation.

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