Polynomials of the Fifth Degree and AboveDate: 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 group associated with it. Start with the rational numbers and adjoin 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/ |
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