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# 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 expansion terminates after the $n-1^{{\text{th}}}$ term. Also, the $n^{{\text{th}}}$ remainder of the Taylor series vanishes; i.e., $\displaystyle R_{n}(x)=\frac{f^{{(n)}}(y)}{n!}x^{n}=0$. Thus, the function is equal to its Taylor series. Hence,

$\begin{array}[]{rl}f(x)&\displaystyle=\sum_{{k=0}}^{{n-1}}\frac{f^{{(k)}}(a)}{% k!}(x-a)^{k}\\ &\\ &\displaystyle=f(a)+\sum_{{k=1}}^{{n-1}}\frac{f^{{(k)}}(a)}{k!}(x-a)^{k}\\ &\\ &\displaystyle=f(a)+(x-a)\sum_{{k=1}}^{{n-1}}\frac{f^{{(k)}}(a)}{k!}(x-a)^{{k-% 1}}\\ &\\ &\displaystyle=f(a)+(x-a)\sum_{{k=0}}^{{n-2}}\frac{f^{{(k+1)}}(a)}{(k+1)!}(x-a% )^{k}.\end{array}$

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 $f(a)=0$.

## Mathematics Subject Classification

12D05*no label found*12D10

*no label found*

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