proof of factor theorem

Suppose that f(x) is a polynomialPlanetmathPlanetmath with real or complex coefficients of degree n-1. Since f is a polynomial, it is infinitely differentiableMathworldPlanetmathPlanetmath. Therefore, f has a Taylor expansionMathworldPlanetmath about a. Since f(n)(x)=0, the terminates after the n-1th term. Also, the nth remainder of the Taylor series vanishes; i.e. (, Rn(x)=f(n)(y)n!xn=0. Thus, the function is equal to its Taylor series. Hence,


If f(a)=0, then f(x)=(x-a)k=0n-2f(k+1)(a)(k+1)!(x-a)k. Thus, f(x)=(x-a)g(x), where g(x) is the polynomial k=0n-2f(k+1)(a)(k+1)!(x-a)k. Hence, x-a is a factor of f(x).

Conversely, if x-a is a factor of f(x), then f(x)=(x-a)g(x) for some polynomial g(x). Hence, f(a)=(a-a)g(a)=0.

It follows that x-a is a factor of f(x) if and only if f(a)=0.

Title proof of factor theorem
Canonical name ProofOfFactorTheorem
Date of creation 2013-03-22 12:39:54
Last modified on 2013-03-22 12:39:54
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 8
Author Wkbj79 (1863)
Entry type Proof
Classification msc 12D05
Classification msc 12D10