# Solution to Sextic Polynomial with Two Real Roots

I have the polynomial

$$f(x;a)=3ax^6+6x^5-9ax^4-4x^3+9ax^2+6x-3a$$

where the variable $a$ is a random variable from the uniform distribution in the range $[0,1)$. When I analyze this function using Mathematica and specific values for $a$, I find there are four complex roots, a negative real root, and a positive real root - I need to extract the positive real root to get an equation $x=x^+(a)$.

I have tried both of Kulkarni's methods${}^{1,2}$ in which he suggests splitting the sextic equation into a quadratic and a quartic or a cubic and another cubic, but numerics in Mathematica show this cannot be done.

At this point, I'm leaning toward using the Kampe de Feriet functions - these are referenced in many places but never given explicitly.

Could someone provide a hint to a solution, or a reference where I could find the application of the Kampe de Feriet function to sextic equations?

Here's what the plot looks like near the region I'm interested in, for $a\in[0.1,0.9]$:

• The only helpful thing I can see is that the polynomial can be written as $3a(x^{2}-1)^{3}+6x^{5}-4x^{3}+6x$. I can't see how to take this any further, though (and I'm sure you'd have noticed if one of your roots was always $\pm1$). – preferred_anon Jan 14 '16 at 22:34
• @DanielLittlewood: His sextic, in fact, had a solvable Galois group. I also made an answer. – Tito Piezas III Mar 23 '16 at 21:14

Actually, your sextic has a solvable Galois group so can be exactly expressed in terms of radicals. The solution turns out to be quite simple. Given

$$3a x ^6 + 6x^5 - 9a x^4 - 4x^3 + 9a x^2 + 6x - 3a = 0\tag1$$

Then the two real roots are,

$$x = u\pm\sqrt{u^2+1}\tag2$$

where,

$$u=\frac{1}{3a}\Big(z_1^{1/3}+\frac{1}{z_1^{1/3}}-1\Big)\tag3$$

and $z_1$ is any non-zero root of,

$$z^2+(9a^2+2)z+1=0\tag4$$

Your sextic's positive real root $x$ is the positive case $\pm$ of $(2)$.

Note: Be careful when testing this numerically, especially using $(3)$. What I find supremely annoying about my version of Mathematica is it gives a complex value to $z^{1/3}$ when $z$ is a negative real number.

P.S.: However, an easy way to verify this is to use resultants and the Mathematica syntax,

Factor[Resultant[x^2 - 2u x - 1, a u^3 + u^2 + 1/3, u]]

which then yields $(1)$, showing that your sextic in just a quadratic $(2)$ in $x$, with coefficients that is cubic $(3)$ in $u$.

After some thought I've decided to use a root-finding method (Newton's method) since I'm working with a polynomial. It can be factored into the form

$$f(x;a) = -3a + x(6 + x(9a + x(-4 + x(-9a + x(6 + 3ax)))))$$

which is most efficient computationally. Ideally I'd also like to know the theoretical expression, but it's unlikely it would be faster to compute than a simple polynomial.

Again, if anyone can point me to literature explaining Kampe de Feriet functions and their applications to polynomial roots, I'd be thankful.