# fundamental theorem of algebra result

This leads to the following theorem:

Given a polynomial^{} $p(x)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\mathrm{\dots}+{a}_{1}x+{a}_{0}$ of degree $n\ge 1$ where ${a}_{i}\in \u2102$, there are exactly $n$ roots in $\u2102$ 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 |