$ \scriptstyle f(x)\ =\ x^n+a_{n-1}x^{n-1}+\ldots+a_0\ =\ \in\ Z[x],\ (x-r)\mid f(x),\ r=\frac{p}{q},\ p,q\in\mathbb{Z},\ q\neq0 $
$ \scriptstyle\Rightarrow\ f(x)\ =\ (x-\frac{p}{q})(x^{n-1}+b_{n-2}x^{n-2}+\ldots+b_0),\ \ b_{n-2},\ldots,b_0\in\mathbb{Q} $
$ \scriptstyle f(r)\ =\ f(\frac{p}{q})\ =\ (\frac{p}{q}-\frac{p}{q})(\textstyle\cdots\cdots\scriptstyle)\ =\ 0 $.
$ \scriptstyle\Rightarrow\ (\frac{p}{q})^n+a_{n-1}(\frac{p}{q})^{n-1}+\ldots+a_1(\frac{p}{q})+a_0\ =\ 0 $.
Now, multiply both sides by $ \scriptstyle q^n $:
$ \scriptstyle p^n+a_{n-1}p^{n-1}q+\ldots+a_1pq^{n-1}+a_0q^n\ =\ 0 $
$ \scriptstyle\Rightarrow\ p^n\ +\ q\cdot(a_{n-1}p^{n-1}+a_{n-2}p^{n-2}q+\ldots+a_1pq^{n-2}+a_0q^{n-1})\ =\ 0 $
$ \scriptstyle\Rightarrow\ p^n\ =\ q\cdot(-a_{n-1}p^{n-1}-a_{n-2}p^{n-2}q-\ldots-a_1pq^{n-2}-a_0q^{n-1}) $
$ \scriptstyle\Rightarrow\ q\mid p^n\ \Rightarrow\ q\mid p $
$ \scriptstyle\Rightarrow\ p\ =\ qs,\ s\in\mathbb{Z}\ \Rightarrow\ r\ =\ \frac{p}{q}\ =\ \frac{qs}{q}\ =\ s $.
$ \scriptstyle\Rightarrow\ r\ \in\ \mathbb{Z} $. $ \scriptstyle\Box $
- --Nick Rupley 03:37, 8 April 2009 (UTC)
