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Why is it so hard to find the roots of polynomial equations?

For polynomials (with real coefficients), in degrees 2, 3, 4, there are the quadratic, cubic, and quartic formula, though the quartic formula is extremely long, so what makes degree 5 special that makes writing down a formula impossible?

Community
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user59677
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    As far as I know, the answer to this rests on Galois Theory, which requires a lot of abstract algebra in order to make sense. Alternatively, the Abel-Ruffini theorem, although I get the sense that Galois theory is more frequently used as justification (I might be wrong on this). At any rate, I don't know enough of that algebra to try at an intuitive explanation, which is why I'm leaving this as a comment. – lamb_da_calculus Jan 25 '13 at 20:52
  • It is possible to express the roots of a quintic polynomial with real coefficients in terms of the coefficients, using just the operations of addition, subtraction, multiplication, division and the taking of $ n $-th roots, if and only if the Galois group of the polynomial is solvable. There are quintic polynomials with Galois group $ S_{5} $, which is not a solvable group. – Haskell Curry Jan 25 '13 at 20:59
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    Dear user, You might like to read this answer, which tries to give an intuitive answer to your question. Regards, – Matt E Jan 25 '13 at 21:00

4 Answers4

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There certainly can be (and is) a formula. It's just not a formula with radicals. Others have mentioned why you can't do it with radicals, having to do with the solvable symmetries, but don't confuse this as to there being no formula at all. A good start might be to investigate Bring Radicals.

ex0du5
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There are some special types quintics which you can solve in radicals (5th and lower roots), but there are some other special ones whose roots have a strange type of symmetry that the expressions we form with radicals can't capture.

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    The paper Solving solvable quintics by D. S. Dummit, in Mathematics of computation, vol 57, 195, July 1991, pp 387-401, explains precisely which quintics are solvable for radicals, and provides a formula in this case. – Andrés E. Caicedo Jan 25 '13 at 21:37
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Without getting into details, the reason five is special (i.e. that there is no formula for the roots in terms of the coefficients, the four arithmetic operations and radicals) is that the group $A_5$ of all even permutations of 5 letters is the smallest non-abelian simple group.

Lior B-S
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I don't think it's so much that $5$ is special but that $2,3$ and $4$ are exceptional. For an irreducible polynomial $p(x) \in \mathbb Q[x]$ there's a general expression for the roots of $p(x)$ in terms of radicals if and only if $\deg p(x) <5$. Now what exactly makes $2,3$ and $4$ different from every other numbers? Well we need some Galois theory and some group theory. To an irreducible polynomial you can associate a group $G$ called it's Galois Group. As it turns out a polynomial is solvable by radicals if and only if its Galois group is solvable. Now because of some elementary group actions arguments it turns out that a $G$ is a subgroup of $S_{\deg p(x)}$ that is the symmetric group on $\deg p(x)$ elements. So the exceptional part is that $S_n$ is only solvable when $n<5$.

That there are polynomials with Galois group $S_n$ is also a somewhat non-trivial fact. Essentially at low degrees nothing can go wrong with solvability, but at higher degrees everything falls apart.

JSchlather
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  • What do you mean that for a polynomial $p$ there's a general formula? – Lior B-S Jan 25 '13 at 21:46
  • @LiorB-S I meant in terms of radicals. I slightly edited it. – JSchlather Jan 25 '13 at 21:49
  • But still, if $p=x^5-2$, the it is irreducible, of degree $5$ but yet it has a root formula, namely the roots are $\zeta\sqrt[5]{2}$, where $\zeta$ runs over 5th roots on unity. – Lior B-S Jan 25 '13 at 21:53
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    @LiorB-S, Yes that's why I said a general formula. There's no one formula in terms of radicals that will work for every $5^{\mathrm{th}}$ degree polynomial. Such as there is in the quadratic case. – JSchlather Jan 25 '13 at 21:57
  • So your formulation is misleading (at least it misled me) because it seems that you fixed p at the beginning and then say there is no formula. – Lior B-S Jan 26 '13 at 07:55