proof of rational root theorem


Let p(x)[x]. Let n be a positive integer with degp(x)=n. Let c0,,cn such that p(x)=cnxn+cn-1xn-1++c1x+c0.

Let a,b with gcd(a,b)=1 and b>0 such that ab is a root of p(x). Then

0=p(ab)=cn(ab)n+cn-1(ab)n-1++c1ab+c0=cnanbn+cn-1an-1bn-1++c1ab+c0.

Multiplying through by bn and rearranging yields:

cnan+cn-1an-1b++c1abn-1+c0bn=0c0bn=-cnan-cn-1an-1b--c1abn-1c0bn=a(-cnan-1-cn-1an-2b--c1bn-1)

Thus, a|c0bn and, by hypothesis, gcd(a,b)=1. This implies that a|c0.

Similarly:

cnan+cn-1an-1b++c1abn-1+c0bn=0cnan=-cn-1an-1b--c1abn-1-c0bncnan=b(-cn-1an-1--c1abn-1-c0bn-1)

Therefore, b|cnan and b|cn.

Title proof of rational root theorem
Canonical name ProofOfRationalRootTheorem
Date of creation 2013-03-22 13:03:53
Last modified on 2013-03-22 13:03:53
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 11
Author Wkbj79 (1863)
Entry type Proof
Classification msc 12D05
Classification msc 12D10