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

The question that follows was inspired by this question:

When trying to solve for the roots of a polynomial equation, the quadratic formula is much more simple than the cubic formula and the cubic formula is much more simple than the quartic formula.

1. That the general solutions to various polynomial equations are so complex and difficult to derive seems to suggest a fundamental limitation in the problem solving capabilities of the mathematical machinery. Does this intuition of mine make any sense? What should I make of it?
2. Why is it that with each successive degree in a polynomial equation, the solution becomes so much more complex? Can I gain some intuition about what makes finding the roots so hard?
3. According to the Abel-Ruffini theorem: "there is no general algebraic solution—that is, solution in radicals— to polynomial equations of degree five or higher." What is so special about the quintic that makes it the cut-off for finding a general algebraic solution?
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I think this should be closed - it's mentioned twice in the question you link to that there is no formula for polynomials of degree > 4, so asking how fast the complexity of the formulas grows is meaningless – BlueRaja - Danny Pflughoeft Jul 27 '10 at 19:33
it's not so hard to find roots numerically... – Jason S Jul 27 '10 at 19:33
possible duplicate of Is there a general formula for solving 4th degree equations? – Akhil Mathew Jul 27 '10 at 19:37
The article you linked to in 3 answers your question 3, and possibly your question 1 if I am understanding it correctly. If there is something about the proof you don't understand, that should be part of your question. Right now I don't really know what you're asking. – Larry Wang Jul 27 '10 at 19:54
@Kaestur: the question is about gaining an intuition. Intuition about why this problem is so hard is very different than a proof that this problem is hard. I recognize that one possible answer to my question might be: it is impossible to gain any intuition about why quintic is unique, all we know how to do is prove it. I'm just curious to see if anyone can surprise me with a particularly elegant way of thinking about this problem. – Ami Jul 27 '10 at 19:59

## 4 Answers

The idea is basically:

Any monic polynomial can be factored as $f(x) = \prod (x - a_i)$, where $a_{1,\dots,n}$ are the roots of the polynomial.

Now if we expand such a product:

$(x - a_1)(x - a_2) = x^2 - (a_1 + a_2)x + a_1a_2$ $(x - a_1)(x - a_2)(x - a_3) = x^3 - (a_1 + a_2 + a_3)x^2 + (a_1a_2 + a_1a_3 + a_2a_3)x - a_1a_2a_3$

And so on. The pattern should be clear.

This means that finding the roots of a polynomial is in fact equivalent to solving systems like the following:

For a quadratic polynomial $x^2 - px + q$, find $a_1,a_2$, such that

$p = a_1 + a_2$

$q = a_1a_2$

For a cubic polynomial $x^3 - px^2 + qx - r$, find $a_1,a_2,a_3$, such that

$p = a_1 + a_2 + a_3$

$q = a_1a_2 + a_1a_3 + a_2a_3$

$r = a_1 a_2 a_3$

And similarly for higher degree polynomials.

Not surprisingly, the amount of "unfolding" that needs to be done to solve the quadratic system is much less than the amount of "unfolding" needed for the cubic system.

The reason why polynomials of degree 5 or higher are not solvable by radicals, can be thought of as: The structure (symmetries) of the system for such a polynomial just doesn't match any of the structures that can be obtained by combining the structures of the elementary operations (adding subtracting, multiplication, division, and taking roots).

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So solving a polynomial can be thought of as solving systems of equations? – Simple Art Dec 26 '15 at 13:52

When you try to solve a degree $n$ equation, there are $n$ roots you have to find (in principle) and none of them is favoured over any of the others, which (in some metaphorical sense) means that you have to break an $n$-fold symmetry in order to write down the roots.

Now the symmetry group of the n roots becomes more and more complicated the larger $n$ is. For $n = 2$, it is abelian (and very small!); for $n = 3$ and $4$ it is still solvable (in the technical sense of group theory), which explains the existence of an explicit formula involving radicals (this is due to Galois, and is a part of so-called Galois theory); for $n = 5$ or more this group is non-solvable (in the technical sense of group theory), and this corresponds to the fact that there is no explicit formula involving radicals.

Summary: The complexity of the symmetry group of the $n$ roots leads to a corresponding complexity in explicitly solving the equation.

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For a different take, some practical problems are discussed in Wilkinson's classic article The Perfidious Polynomial. If you can't access it, check what Wikipedia has to say on the subject.

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Jim Wilkinson's prize-winning article can be read here; briefly, one frequent source of trouble in numerical polynomial root finding is our habit of expressing polynomials in the monomial basis, and it happens that there are polynomials like $\prod_i (x-i)$ that numerically behave very poorly in rootfinding when expressed in the monomial basis. – J. M. Apr 20 '11 at 8:51
@Ｊ.Ｍ. This link is broken. – I. J. Kennedy Oct 25 '14 at 23:48
I think I found it. maa.org/sites/default/files/pdf/upload_library/22/Chauvenet/… – Neil Apr 6 '15 at 22:07
@I.J.Kennedy, Neil has given you a working link; thank him. – J. M. May 1 '15 at 13:42

As an introduction to this important issue in mathematics, I shall first consider the whole issue as simplified polynomial or (general trinomial equation) by providing only and absolutely one correct root (absolutely convergent to algebraic real irrational number always) to the following trinomial equation (without using any approximation methods), say for $n$-th degree polynomial equation $$x^n + x^m = 1,$$ where $m < n$ are two distinct positive integers, the real root is given by the following power series:

\begin{align} x &= 1-\frac{1}{n}+ \sum_{i=2}^\infty \frac{(-1)^i \prod_{j=1}^{i-1}(im-jn+1)}{i!n^i} \\ &= 1-\frac{1}{n}+\frac{2m-n+1}{2!n^2}- \frac{(3m-2n+1)(3m-n+1)}{3!n^3} +\frac{(4m-3n+1)(4m-2n+1)(4m-n+1)}{4!n^4} -\frac{(5m-n+1)(5m-2n+1)(5m-3n+1)(5m-4n+1)}{5!n^5} +\dotsb \end{align}

In so many cases the root can be radical form, but the big question is that always can be converted to radical form, for a more useful reference see the link below:

Reference: (1994), For a reference to more general solution of this Trinomial equation $$ax^n +bx^m +c = 0$$ click, OK, then click National then type in the search button "solution of equations by power series" at this link http://opac.nl.gov.jo/uhtbin/cgisirsi

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I don't think this answers the question. I also think it's odd that this is identical to another recent answer of yours. – pjs36 Nov 26 '15 at 16:12
I thought that I could provide a little insight to the problem even repeatedly, I would also appreciate the downvoters if they prove me wrong openly with their real identities, I know also many have deleted my same answer for several times without any obvious reason, it is quite simple, just assume any arbitrary example even with radical known roots and see then if my answer is wrong, mentioning that what is given in the answer is very little from the original reference that I strictly believe is very useful subject, I can provide infinitely many radical solution from the provided series – bassam karzeddin Nov 26 '15 at 16:30
It seems that for more than ten times nobody could justify clearly their downvotes or denying the formula openly,or even adding any information, plus blocking me from asking questions, so this was subjected to be deleted always, wondering if maths becoming so boring up to this limit, It is so simple, it is an identity, (One equation & One unknown principle), it has so many applications, imagine if it is coming from a reputable Journal (with a very long proof) how the impact on the mathematicians would be, however, deleting the FORMULA REPEATEDLY wouldn't in any case invalidate it, isn't it? – bassam karzeddin Dec 3 '15 at 7:33