# 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 $n-{1}^{\text{th}}$ term. Also, the ${n}^{\text{th}}$ remainder of the Taylor series vanishes; i.e. (http://planetmath.org/Ie), ${R}_{n}(x)={\displaystyle \frac{{f}^{(n)}(y)}{n!}}{x}^{n}=0$. Thus, the function is equal to its Taylor series. Hence,

$\begin{array}{cc}\hfill f(x)& ={\displaystyle \sum _{k=0}^{n-1}}{\displaystyle \frac{{f}^{(k)}(a)}{k!}}{(x-a)}^{k}\hfill \\ & \\ & =f(a)+{\displaystyle \sum _{k=1}^{n-1}}{\displaystyle \frac{{f}^{(k)}(a)}{k!}}{(x-a)}^{k}\hfill \\ & \\ & =f(a)+(x-a){\displaystyle \sum _{k=1}^{n-1}}{\displaystyle \frac{{f}^{(k)}(a)}{k!}}{(x-a)}^{k-1}\hfill \\ & \\ & =f(a)+(x-a){\displaystyle \sum _{k=0}^{n-2}}{\displaystyle \frac{{f}^{(k+1)}(a)}{(k+1)!}}{(x-a)}^{k}.\hfill \end{array}$

If $f(a)=0$, then $f(x)=(x-a){\displaystyle \sum _{k=0}^{n-2}}{\displaystyle \frac{{f}^{(k+1)}(a)}{(k+1)!}}{(x-a)}^{k}$. Thus, $f(x)=(x-a)g(x)$, where $g(x)$ is the polynomial $\sum _{k=0}^{n-2}}{\displaystyle \frac{{f}^{(k+1)}(a)}{(k+1)!}}{(x-a)}^{k$. 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 |