ProofOfFactorTheorem 2002-05-25 20:37:59 2008-09-08 19:09:15 Proof factor theorem 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here 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 \PMlinkescapetext{expansion} terminates after the $n-1^{\text{th}}$ term. Also, the $n^{\text{th}}$ remainder of the Taylor series vanishes; \PMlinkname{i.e.}{Ie}, $\displaystyle R_n(x)=\frac{f^{(n)}(y)}{n!}x^n=0$. Thus, the function is equal to its Taylor series. Hence, \begin{center} $\begin{array}{rl} f(x) & \displaystyle =\sum_{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a)^k \\ & \\ & \displaystyle =f(a)+\sum_{k=1}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a)^k \\ & \\ & \displaystyle =f(a)+(x-a)\sum_{k=1}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a)^{k-1} \\ & \\ & \displaystyle =f(a)+(x-a)\sum_{k=0}^{n-2}\frac{f^{(k+1)}(a)}{(k+1)!}(x-a)^k. \end{array}$ \end{center} If $f(a)=0$, then $\displaystyle f(x)=(x-a)\sum_{k=0}^{n-2}\frac{f^{(k+1)}(a)}{(k+1)!}(x-a)^k$. Thus, $f(x)=(x-a)g(x)$, where $g(x)$ is the polynomial $\displaystyle \sum_{k=0}^{n-2}\frac{f^{(k+1)}(a)}{(k+1)!}(x-a)^k$. 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 $f(a)=0$.