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### Polynomials of the Fifth Degree and Above

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Date: 07/28/2001 at 16:45:51
From: Peter
Subject: Polynomials

Dear Dr. Math,

I know how to find the root of a polynomial of the second degree: I
mean a polynomial of the form: ax^2 + bx + c = 0. But what about a
polynomial of the third degree? How can we find the root? And what
about the other degrees? If there is a method, please I would like to
know the proof of it.

One more question:
I have read in a book that a mathematician (I don't remember his
name), has proved that there is no method to resolve polynomials of
the fifth degree and above. How did he proved that?

Thanks a lot,
Peter.
```

```
Date: 07/30/2001 at 14:50:21
From: Doctor Rob
Subject: Re: Polynomials

Thanks for writing to Ask Dr. Math, Peter.

For solutions to the cubic (third degree) and quartic (fourth degree)
equations, see the following page from our Frequently Asked
Questions (FAQ):

http://mathforum.org/dr.math/faq/faq.cubic.equations.html   .

It is true that in 1824 Niels Henrick Abel proved that it is not
possible to derive a solution to a general irreducible quintic (fifth
degree) equation from its coefficients by a finite sequence of
additions, subtractions, multiplications, divisions, and taking n-th
roots. Evariste Galois extended this to all higher degrees. The tool
Galois used is something called Group Theory.

For a given polynomial p(x) with rational coefficients, you can find a
to them all the roots of p(x), that is, solutions of p(x) = 0. Extend
linearly to form a field. Then the Galois group of p(x) consists of
all the automorphisms of this field that leave the rational numbers
fixed and permute the roots of p(x). The group operation is
composition of automorphisms. It turns out that the equation p(x) = 0
is solvable with radicals if and only if the Galois group G of p(x) is
a solvable group. That means that there is a sequence of normal
subgroups

G = G(0) > G(1) > G(2) > ... > G(n) = {1}

such that for each i with 0 <= i < n, G(i)/G(i+1) is a cyclic group.
Furthermore, it turns out that almost all (in some technical sense)
polynomials of degree k with rational coefficients have Galois group
isomorphic to S(k), the group of permutations of k letters, and S(k)
is not a solvable group for k > 4.

Some quintic and higher degree irreducible polynomial equations can
be solved using radicals. A trivial example is x^k - 2 = 0, one of
whose solutions is x = 2^(1/k), the k-th root of 2. There are other,
less trivial examples of this, such as x^5 - 5*x + 12 = 0.

This is fairly advanced mathematics in the field of Abstract Algebra,
which is generally not studied until the first year of graduate
school.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Modern Algebra
High School History/Biography
High School Polynomials

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