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

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

• Define "exactly solved" – Balarka Sen Jun 16 '14 at 11:17
• @BalarkaSen Not approximate or numerical solution. Is solution by radicals an all-encompassing, universal concept of exact solvability? – Nikos M. Jun 16 '14 at 11:26
• Try this then. – Balarka Sen Jun 16 '14 at 11:30
• @BalarkaSen, nice it seems there are other methods, however from a quick look it seems this is only for general quintics, no? – Nikos M. Jun 16 '14 at 11:32
• Yes. In general, higher degree polynomials are not solvable by these tricks. It is provably true for at least sextics : they are not solvable by one-variable holomorphic functions. – Balarka Sen Jun 16 '14 at 12:39

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.

• +1, i really like the prologue about the symbolic meaning of "radicals" and how these can be extended to new "forms" that indeed can "solve" sth beyond. (if i could give 2 points, i would give it) – Nikos M. Jun 16 '14 at 11:42
• a question, how is this "theory" or "field of mathematics" called (involving "Bring radical" etc..)? – Nikos M. Jun 16 '14 at 11:43
• @NikosM. Which particular field do you mean? If you're interested in questions like "can we reduce solving $n$ degree polynomials to solving particular finite-dimensional familes of polynomials", then I imagine Galois theory is the place to start. (Also, it's generally not advised to accept answers too quickly, to give others a chance to answer). – Jack M Jun 16 '14 at 11:45
• So the theory involving Bring radicals is still (an extension of?) Galois Theory? – Nikos M. Jun 16 '14 at 11:46
• @NikosM. I've never studied Bring radicals, so I couldn't say. My suggestion would be to look for a proof that all quintics can be solved using Bring radicals, and see what kind of math is used. – Jack M Jun 16 '14 at 11:48

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.

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.

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.

• great, i would prefer some elaboration on any alternative methods or formalisms for solvability of (arbitrary) polynomial equations of deegre $\ge 5$, if any? – Nikos M. Jun 16 '14 at 11:36