Timeless Theorems of Mathematics/Rational Root Theorem

The rational root theorem states that, if a rational number $\frac{p}{q}$ (where $p$ and $q$ are relatively prime) is a root of a polynomial with integer coefficients, then $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. In other words, for the polynomial, $P(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+a_{n-2}{x}^{n-2}+...+a_0$, if $P\left(\frac{p}{q}\right)=0$, (where $a_i\in \mathbb{Z}$ and $a_0,a_n\ne 0$) then $\frac{a_0}{p}\in \mathbb{Z}$ and $\frac{a_n}{q}\in \mathbb{Z}$

Proof

Let $P(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+a_{n-2}{x}^{n-2}+...+a_0$, where $a_i\in \mathbb{Z}$.

Assume $P(\frac{p}{q})=0$ for coprime $p,q\in \mathbb{Z}$. Therefore, $P(\frac{p}{q})=a_n(\frac{p}{q}{)}^n+a_{n-1}(\frac{p}{q}{)}^{n-1}+a_{n-2}(\frac{p}{q}{)}^{n-2}+...+a_1(\frac{p}{q})+a_0=0$ $\Rightarrow a_n{p}^n+a_{n-1}{p}^{n-1}q+a_{n-2}{p}^{n-2}{q}^2+...+a_{1}p{q}^{n-1}+a_0{q}^n=0$ $\Rightarrow p(a_n{p}^n-1+a_{n-1}{p}^{n-2}q+a_{n-3}{p}^{n-2}{q}^2+...+a_1{q}^{n-1})=-a_0{q}^n$

Let $w=a_n{p}^n-1+a_{n-1}{p}^{n-2}q+a_{n-3}{p}^{n-2}{q}^2+...+a_1{q}^{n-1}\in \mathbb{Z}$

Thus, $w=-\frac{a_0{q}^n}{p}$

As $p$ is coprime to $q$ and $w\in \mathbb{Z}.$, thus $\frac{a_0}{p}\in \mathbb{Z}$.

Again, $a_n{p}^n+a_{n-1}{p}^{n-1}q+a_{n-2}{p}^{n-2}{q}^2+...+a_{1}p{q}^{n-1}+a_0{q}^n=0$ $\Rightarrow q(a_{n-1}{p}^{n-1}+a_{n-2}{p}^{n-2}q+...+a_{1}p{q}^{n-2}+a_0{q}^{n-1})=-a_n{p}^n$

Let $(a_{n-1}{p}^{n-1}+a_{n-2}{p}^{n-2}q+...+a_{1}p{q}^{n-2}+a_0{q}^{n-1})=v\in \mathbb{Z}$

Thus, $qv=-a_n{p}^n$ $\Rightarrow v=-\frac{a_n{p}^n}{q}$

As $q$ is coprime to $p$ and $v\in \mathbb{Z}.$, thus $\frac{a_n}{p}\in \mathbb{Z}$.

$\therefore$ For $P(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+a_{n-2}{x}^{n-2}+...+a_0$, if $P(\frac{p}{q})=0$, (where $a_i\in \mathbb{Z}$ and $a_0,a_n\ne 0$) then $\frac{(a_0)}{p}\in \mathbb{Z}$ and $\frac{a_n}{q}\in \mathbb{Z}$. [Proved]