# Polynomial equations of degree larger than 4

It is accepted that there are no general solutions for polynomial equations of degree higher than 4, unless they have some unique features. We know that if we could factor the polynomial into polynomial of degrees smaller than 5, we may find the roots for each such polynomial with degree smaller than 5.

Are there systematic tools for deciding whether polynomials with integer coefficients are irreducible over the rationals, and for factoring them over the rationals if they are not irreducible over the rationals?

• There are general solutions for fifth, sixth, and seventh degree equations, they just involve functions that can't be expressed in terms of radicals. – Matt Samuel Aug 22 '15 at 1:31
• Note that asking whether a polynomial can be factored is not the same as asking whether one can find its roots. Anyway, in principle, one can find the Galois group of any given polynomial, and then determine whether that group is a solvable group. If it is, we can solve the polynomial in radicals; if not, not. In practice, it's not so easy to find the Galois group. The question of how to find the Galois group has come up on this website before, so try to have a look around. – Gerry Myerson Aug 22 '15 at 3:13
• As I said, you clearly do not know what you are talking about. First you claimed "there are no general solutions for polynomial equations of degree higher than 4". That is utterly meaningless unless you are talking about expressing the roots using only radicals over the original field! Then you claim that it is impossible to factor some polynomials. Since you said nothing about radicals, this claim is false. My later paragraphs deal with the question about radical field extensions, which is the only situation where your first claim makes any sense at all. – user21820 Aug 22 '15 at 7:01
• Have you tried reading the Wikipedia page? Yes, given a polynomial with rational coefficients, there is a systematic way to decide whether it is irreducible over the rationals, and a systematic way to factor it into polynomials that are irreducible over the rationals. For polynomials of degree 5 or greater, there is, in general, no closed form for their roots in terms of radicals and field operations. By the way, if you want to be certain that I see a comment, you have to put @Gerry into it. – Gerry Myerson Aug 23 '15 at 6:22
• Since the question is on hold, I can't post an answer. What you can do is edit ithe statement of the question so as to clarify (people shouldn't have to wade through a long thread of comments to understand the question), and then post to the meta site, noting that you have edited the question, and asking for it to be reopened. If it gets reopened, then you can post an answer yourself, if you wish, based on what you have learned from the discussion. – Gerry Myerson Aug 24 '15 at 1:34

## 1 Answer

Any polynomial over a field factors completely over its splitting field. For real polynomials, it turns out that adjoining $\sqrt{-1}$ to $\mathbb{R}$ gives the algebraic closure, and so every real polynomial factors over $\mathbb{C}$. So your question's assumption is incorrect.

On the other hand, there are equations using radicals that more or less give the roots of a quadratic, cubic or quartic polynomial over a field of characteristic zero. In some cases these formulae are somewhat silly because it is usually not easier to find the cube-root of a complex number than to find the roots of the original cubic polynomial! It is useless to say that we can use exponentials, trigonometric functions and their inverses, since they also cannot be simplified. If we eventually want to compute an approximation, we do not even want to use the cubic equation! There are root-finding algorithms that work for any arbitrary polynomial over the reals and make formulae using radicals completely redundant in the real world where we do not need infinite precision.

But it is still an interesting question whether there is a general formula for the roots of a quintic polynomial over a field $F$ that only uses radicals, in other words whether the roots are in some radical extension of $F$. It turns out that it is possible exactly when the Galois group of the Galois closure of the field is solvable ("solvable" arose from this very problem of trying to solve polynomials), in other words there is a chain of groups from that group to the trivial group such that the quotient between consecutive groups in the chain is cyclic. So everything reduces to the nature of the Galois group. See https://math.stackexchange.com/a/38901/21820 for an overview of the possible Galois group of an irreducible quintic. I know that for a quartic over the rationals there are easy ways to determine the Galois group just by looking at the coefficients. I do not know if there are deterministic tests for quintics.

• Finding the cube root of a complex number is certainly not nearly as hard as solving a cubic equation. – Matt Samuel Aug 22 '15 at 2:13
• finding cube roots is trivial by de Moivre's theorem: $z^3=re^{it}$ gives $z\in\{\sqrt{r}e^{it/3},\sqrt{r}e^{i(t+2\pi)/3},\sqrt{r}e^{i(t+4\pi)/3}\}$ – oldrinb Aug 22 '15 at 2:16
• @MattSamuel: It is. Tell me the exact value of $\sqrt{1+2i}$ using only arithmetic. – user21820 Aug 22 '15 at 2:17
• No. But I have no problem using trigonometry. The point is there's nothing to solve, it's just plugging something in. – Matt Samuel Aug 22 '15 at 2:19
• @oldrinb: Tell me the exact value of $t$ for $(1+2i)$. It is no point saying you know the cube roots when I too can write $\exp(\frac{1}{3}\ln(1+2i))$ (multi-valued) to express them. – user21820 Aug 22 '15 at 2:19