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,

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