# 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), $\displaystyle R_{n}(x)=\frac{f^{(n)}(y)}{n!}x^{n}=0$. 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 $f(a)=0$.

Title proof of factor theorem ProofOfFactorTheorem 2013-03-22 12:39:54 2013-03-22 12:39:54 Wkbj79 (1863) Wkbj79 (1863) 8 Wkbj79 (1863) Proof msc 12D05 msc 12D10