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 $n-1^{{th}}$ term. Also, the $n^{{th}}$ 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,

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 $f(a)=0$