proof of factor theorem
Suppose that f(x) is a polynomial with real or complex coefficients of degree n-1. Since f is a polynomial, it is infinitely differentiable
. Therefore, f has a Taylor expansion
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. (http://planetmath.org/Ie), Rn(x)=f(n)(y)n!xn=0. Thus, the function is equal to its Taylor series. Hence,
f(x)=n-1∑k=0f(k)(a)k!(x-a)k=f(a)+n-1∑k=1f(k)(a)k!(x-a)k=f(a)+(x-a)n-1∑k=1f(k)(a)k!(x-a)k-1=f(a)+(x-a)n-2∑k=0f(k+1)(a)(k+1)!(x-a)k.
If f(a)=0, then f(x)=(x-a)n-2∑k=0f(k+1)(a)(k+1)!(x-a)k. Thus, f(x)=(x-a)g(x), where g(x) is the polynomial n-2∑k=0f(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 |