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# 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, $x_{1}=x_{2}=x_{3}=x_{4}=0$.

## Mathematics Subject Classification

12D99*no label found*30A99

*no label found*

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## Comments

## grammatical disagreement

The "is" needds to be plularized to "are" in the statement of the theorem.