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'proof of factor theorem'
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| Title of object: |
proof of factor theorem |
| Canonical Name: |
ProofOfFactorTheorem |
| Type: |
Proof |
| Created on: |
2002-05-25 20:37:59 |
| Modified on: |
2006-07-24 08:16:57 |
| Classification: |
msc:12D05, msc:12D10 |
Preamble:
% 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 |
Content:
Suppose that $f(x)$ is a polynomial 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 $f(a)=0$ we have $\displaystyle f(x)=(x-a)\sum_{k=0}^{n-2}\frac{f^{(k+1)}(a)}{(k+1)!}(x-a)^k$. 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 $f(a)=0$. |
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