Date: 03/12/2001 at 17:45:53 From: Joan Subject: Number theory Please explain LaGrange's Theorem on the number of roots of a polynomial. Thank you.
Date: 03/13/2001 at 10:35:36 From: Doctor Rob Subject: Re: Number theory Thanks for writing to Ask Dr. Math, Joan. This theorem states that a polynomial of degree n over a field has at most n roots in that field. Usually this is proved by induction on n. Let f(x) be the polynomial. If n = 0, then f(x) is a nonzero constant, and so has no roots, so the statement is true for n = 0. Now suppose that n > 0, and that the statement is true for all polynomials of degree n - 1. Then f(x) may have no roots in the field, in which case the theorem holds. If it has roots in the field, let r be one of them. Then divide f(x) by x - r with remainder: f(x) = (x-r)*g(x) + c, c a constant. Evaluate this at x = r: 0 = f(r) = (r-r)*g(r) + c = c, since r is a root of f(x). Since c = 0, that means that f(x) = (x-r)*g(x). Suppose that s is any root of f(x) in the field. Then 0 = (s-r)*g(s). This implies, since we are working in a field, that either s - r = 0, or g(s) = 0 (or both). Thus either s = r, or s is a root of g(x), or both. Now the induction hypothesis applies to g(x) whose degree is n - 1. I leave the rest of the proof to you. By the way, this is false if you are not working over a field. The polynomial x^2 - x has four roots in the ring of integers modulo 6, namely 0, 1, 3, and 4. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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