How to solve a 6th-degree polynomial

Question

How would you solve a polynomial with the equation $$ax^6 + bx^5 + cx^4 + dx^3 + ex^2 + fx + g = 0, \text{ ?}$$ with all of the coefficients being positive except for the constant.

Answer 1

If I knew the values of the coefficients I would try to solve it by methods relying heavily on numerical computation, quite likely Newton's method. There might not be any real solutions, but a famous theorem assures us that there would be complex solutions. Another famous theorem says the solutions are not functions of the coefficients that can be built using addition, subtraction, multiplication, division, and the extraction of square roots, cube roots, fourth roots, etc. That can be done for $4$th-degree and lower-degree polynomials, as when one solves a quadratic equation via the usual formula. In some special cases it can be done for higher-degree polynomials, but we would have to know something about the coefficients to know that we had such a special case.

How do you study the roots as functions of the coefficients? I'd might start by googling "algebraic function" and "complex variables" or "complex analysis". There are some theorems about those, and I suspect a lot more is known than the basic theorems about them that I've seen.

Maybe there are also some clever geometric methods.

Answer 2

The root of polynomial could be irrational, and in practice, you normally can only get approximation of them through numerical methods.

The equation you have above is completely general and there is no additional insights on what could be leveraged. Your best chance is pick a random starting point and then use the Bairstow's method. (That will, at the end, give you all roots including the complex ones)

https://en.wikipedia.org/wiki/Bairstow%27s_method

Note that as Bairstow's method depends on the initial starting point, it could be the case that unluckily the algorithm converge, just try different starting point in that case.

Answer 3

The sextic does not usually have a solution that can be expressed in terms of finitely many algebraic operations (adding, subtracting, multiplying, dividing and taking roots). This is a result proved by Abel (and Galois), which in fact holds for any polynomial of degree $5$ or greater.

What this means is that there is no general way to analytically obtain the roots of these types of polynomials, although one can numerically approximate solutions.

Note that this doesn't mean that we can never solve quintics or higher degree polynomials by hand, for example it doesn't take too much effort to see that $$ x^6 -1 $$ has roots $-1$ and $1$. What it does mean is that if you have some arbitrarily chosen $a, b, \dotsc, g$ then you will most likely not be able to explicitly calculate the roots of $ax^6 + \dotsb + g$ exactly.

Answer 4

Mathworld claims that the general sextic can be solved using Kampé de Fériet functions. Mind you, this is not something one would do in practice...