fundamental theorem of algebra result

This leads to the following theorem:

Given a polynomial $p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots+a_{1}x+a_{0}$ of degree $n\geq 1$ where $a_{i}\in\mathbb{C}$ , there are exactly $n$ roots in $\mathbb{C}$ to the equation $p(x)=0$ if we count multiple roots.

Proof The non-constant polynomial $a_{1}x-a_{0}$ has one root, $x=a_{0}/a_{1}$ . Next, assume that a polynomial of degree $n-1$ has $n-1$ roots.

The polynomial of degree $n$ has then by the fundamental theorem of algebra a root $z_{n}$ . With polynomial division we find the unique polynomial $q(x)$ such that $p(x)=(x-z_{n})q(x)$ . The original equation has then $1+(n-1)=n$ roots. By induction, every non-constant polynomial of degree $n$ has exactly $n$ roots.

For example, $x^{4}=0$ has four roots, $x_{1}=x_{2}=x_{3}=x_{4}=0$ .

Title	fundamental theorem of algebra result
Canonical name	FundamentalTheoremOfAlgebraResult
Date of creation	2013-03-22 14:22:01
Last modified on	2013-03-22 14:22:01
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	7
Author	rspuzio (6075)
Entry type	Theorem
Classification	msc 12D99
Classification	msc 30A99