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 $a_{0},a_{n}\neq 0$) 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, $P({\frac {p}{q}})=a_{n}({\frac {p}{q}})^{n}+a_{n-1}({\frac {p}{q}})^{n-1}+a_{n-2}({\frac {p}{q}})^{n-2}+...+a_{1}({\frac {p}{q}})+a_{0}=0$ $\Rightarrow a_{n}p^{n}+a_{n-1}p^{n-1}q+a_{n-2}p^{n-2}q^{2}+...+a_{1}pq^{n-1}+a_{0}q^{n}=0$ $\Rightarrow p(a_{n}p^{n}-1+a_{n-1}p^{n-2}q+a_{n-3}p^{n-2}q^{2}+...+a_{1}q^{n-1})=-a_{0}q^{n}$

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

Let $(a_{n-1}p^{n-1}+a_{n-2}p^{n-2}q+...+a_{1}pq^{n-2}+a_{0}q^{n-1})=v\in \mathbb {Z}$

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 $a_{0},a_{n}\neq 0$) then ${\frac {(a_{0})}{p}}\in \mathbb {Z}$ and ${\frac {a_{n}}{q}}\in \mathbb {Z}$. [Proved]