# proof of rational root theorem

Let $p(x)\in\mathbb{Z}[x]$. Let $n$ be a positive integer with $\operatorname{deg}p(x)=n$. Let $c_{0},\ldots,c_{n}\in\mathbb{Z}$ such that $p(x)=c_{n}x^{n}+c_{n-1}x^{n-1}+\cdots+c_{1}x+c_{0}$.

Let $a,b\in\mathbb{Z}$ with $\operatorname{gcd}(a,b)=1$ and $b>0$ such that $\displaystyle\frac{a}{b}$ is a root of $p(x)$. Then

$\begin{array}[]{ll}0&\displaystyle=p\left(\frac{a}{b}\right)\\ \\ &\displaystyle=c_{n}\left(\frac{a}{b}\right)^{n}+c_{n-1}\left(\frac{a}{b}% \right)^{n-1}+\cdots+c_{1}\cdot\frac{a}{b}+c_{0}\\ \\ &\displaystyle=c_{n}\cdot\frac{a^{n}}{b^{n}}+c_{n-1}\cdot\frac{a^{n-1}}{b^{n-1% }}+\cdots+c_{1}\cdot\frac{a}{b}+c_{0}.\end{array}$

Multiplying through by $b^{n}$ and rearranging yields:

$\begin{array}[]{rl}c_{n}a^{n}+c_{n-1}a^{n-1}b+\cdots+c_{1}ab^{n-1}+c_{0}b^{n}&% =0\\ \\ c_{0}b^{n}&=-c_{n}a^{n}-c_{n-1}a^{n-1}b-\cdots-c_{1}ab^{n-1}\\ \\ c_{0}b^{n}&=a\left(-c_{n}a^{n-1}-c_{n-1}a^{n-2}b-\cdots-c_{1}b^{n-1}\right)\\ \end{array}$

Thus, $a|c_{0}b^{n}$ and, by hypothesis, $\operatorname{gcd}(a,b)=1$. This implies that $a|c_{0}$.

Similarly:

$\begin{array}[]{rl}c_{n}a^{n}+c_{n-1}a^{n-1}b+\cdots+c_{1}ab^{n-1}+c_{0}b^{n}&% =0\\ \\ c_{n}a^{n}&=-c_{n-1}a^{n-1}b-\cdots-c_{1}ab^{n-1}-c_{0}b^{n}\\ \\ c_{n}a^{n}&=b\left(-c_{n-1}a^{n-1}-\cdots-c_{1}ab^{n-1}-c_{0}b^{n-1}\right)\\ \end{array}$

Therefore, $b|c_{n}a^{n}$ and $b|c_{n}$.

Title proof of rational root theorem ProofOfRationalRootTheorem 2013-03-22 13:03:53 2013-03-22 13:03:53 Wkbj79 (1863) Wkbj79 (1863) 11 Wkbj79 (1863) Proof msc 12D05 msc 12D10