Is a polynomial equation of degree $\ge 5$ not solvable by any way?

Question

By Galois Theory an arbitrary polynomial equation of degree $\ge 5$ is not solvable using radicals, unlike the polynomial equation of second degree which is solvable by radicals (because of the alternating group of order 5, the symmetry group is not soluble).

Is this the only way an arbitrary polynomial equation can be (exactly) solved?

Does Galois Theory provide universal concept of general solvability?

Are there other methods except using radicals (for exact, not approximate or numerical solution)?

PS. A related question

Answer 1

As my Group theory professor told me, giving a root of $x^2-2=0$ the name $\sqrt 2$ doesn't magically "solve" the equation. Calling a number $\sqrt 2$ is just saying "it solves $x^2-2=0$". You haven't gained any new information. Similarly, if $p(x)$ is an irreducible polynomial, then simply writing down $p(x)=0$ could be considered to be "solving" the equation, in the sense that, in a certain technical sense, the fact that $x$ solves that equation contains all relevant information about $x$.

Solving by radicals means that we wish to express roots of a polynomial in terms of the roots of a particular family of polynomials $x^2-a$. One reason to do this is that it means that we can numerically approximate those roots using methods to numerically approximate roots of that simpler family of polynomials. This is a common strategy in mathematics. For example, in trigonometry we try to express all quantities using $\cos$ and $\sin$. In a sense, this doesn't mean we've "calculated" those quantities, since $\cos$ and $\sin$ are just names given to particular ratios of line segments, but it does mean that we can reduce the problem of approximating various quantities to just approximating $\cos$ and $\sin$.

The Bring radical represents the applicaiton of this strategy to quintics. We choose a particular one-dimensional (in the vector space sense) family of polynomials and invent a special symbol to refer to a particular root of a polynomial in that family. It can be shown that in this way we can express the roots of any quintic.

Answer 2

With "solved by radicals" is meant that the root is expressed in closed form using coefficients, using actual numbers for coefficients you might get something like these examples: $$x=3$$ $$x=\frac{\sqrt{13}+6i}{3i}$$ $$x= \frac{1}{6}-\frac{\left(\frac{7}{2}\right)^{2/3} \left(1+i \sqrt{3}\right)}{6 \sqrt[3]{-1+3 i \sqrt{3}}}-\frac{1}{12} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3 i \sqrt{3}\right)}$$

And the Galois fella tells you that there are some polynomials for which such a strange roots exist that cannot be written in form like this. It's not even about the solving (finding them), the roots themselves are such numbers that are impossible to express in this kind of expression.

Does that answer your question? If your question was if these numbers can be written in some other form - yes, why not, but it doesn't mean it will seem more precise.

For me exact form is one that defines the number and "smallest root of ..." can seem more graspable than some of the expressions above.

Answer 3

There are solvable quintics of course, but there is also a general solution of the quintic, using more than radicals, namely, using hypergemetric functions or theta functions.

See for example Quintic equation on MathWorld, and Quintic function, Bring radical on Wikipedia.

For the sextic, a Google search will show you several results, I'll put more references here if needed.

Answer 4

There certainly are quintics that can be solved by radicals, but the point is that there are some whose zeroes can't be expressed in terms of roots of the coefficients.