Trying to find to find the zeroes of a sextic polynomial to write an analytical form:
$$r^6-r^5-r^4 \cdot C+r^2 \cdot D+E=0$$
actually $r=\frac{x}{b}$ and I want to find an analytical form for $x$ like for a quartic function.
I'm guessing I will need the zeroes here but I'm not sure how to solve it.
C,D and E are real numbers generated by different data points so they change for every iteration which is one of the reason I want to have x on analytical from so that I can plot it against the data that generates C,D and E.
If there is no way to get an analytic solution is there anyway of finding the zeroes by a numerical method like newton's method?
Thank you.
This is still the general sextic in disguise, so there is no analytic solution in radicals.
You can use a Tschirnhausen transformation in radicals to simultaneously eliminate the $x^{n-1}$ and $x^{n-3}$ terms of the general $n$th deg equation. The sextic becomes,
$$x^6+an^2x^4+bn^4x^2+cn^5x+dn^6=0\tag1$$
where $n$ is free parameter. Then use the substitution $r = 1/x$ on your sextic to get,
$$Ex^6+Dx^4-Cx^2-x+1=0$$
This is just,
$$x^6+px^4+qx^2+tx-t=0\tag2$$
If you equate the last two terms of $(1)$,
$$cn^5=-dn^6$$
then solve for $n$, you will see that $(1)$ and $(2)$ have the same form, hence your sextic is just the general sextic in disguise, so no solution in radicals.