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It is not generally possible to determine the roots of a polynomial whose grade is bigger than 4 in terms of roots and basic operations. But I heard that it is possible to give a criteria whether a polynomial has such a solution.

For instance Wikipedia tells that the roots of a polynomial of degree $5$ are expressible in terms of roots and basic operations, if it is representable in the form

$$x^5 + \frac{5\mu^4(4\nu + 3)}{\nu^2 + 1}x + \frac{4\mu^5(2\nu + 1)(4\nu + 3)}{\nu^2 + 1} = 0,$$

where $\mu$ and $\nu$ are rational numbers.

Given the case that the polynomials roots are expressible in such a form, is it possible to give an algorithm that computes the solution in this form?

I am not an expert on that topic, just a student who is interested in maths.

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Are you asking the same thing that was asked in ? You may want to have a look at that question... – Steven Stadnicki May 18 '11 at 21:39
up vote 5 down vote accepted

Yes, it's called Galois Theory.

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I read the article and think, that you can certainly show that a polynomial's roots are describable in terms of radicands using the Galois Theory, but it doesn't explains you how to find them. – FUZxxl May 18 '11 at 20:48
@FUZxxl, see for instance… – lhf May 18 '11 at 21:41

While one cannot express the terms of polynomials of degree 5 and higher with basic operations and roots (Abel-Ruffini), it is possible to explicitly calculate the roots of polynomials of degree five and higher if you use other tools.

See this MO question.

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My question was, if the polynomial is expressable in terms of radicals, is it possible to give an algorithm to find them? – FUZxxl May 19 '11 at 7:00

Yes, assuming an expression for a root exists, there is a simple algorithm that will find it: define an ordering on all such expressions (e.g. a lexical ordering), and test each one until you find a root of the input polynomial.

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On the other hand, the problem of deciding whether a particular expression is a root of the given polynomial may not be so easy. I presume that it can be done numerically using some super-exponential error bound in terms of the length of the expression resulting from substituting the candidate root into the polynomial. Or is it guaranteed that it will evaluate to 0 symbolically? I'm really not sure. – Dan Brumleve Jun 2 '11 at 1:24
I'm pretty well convinced now that a symbolic method will work for the test, essentially because of algebraic independence (e.g. the cube root of p will never equal the square root of q). If anyone knows otherwise, I welcome a correction. – Dan Brumleve Jun 2 '11 at 1:40

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