I have found a question where I have to determine the irreducibility of a polynomial $f$ in $\textbf{Q}[X]$. Namely, $$ f=2X^3-X-6.$$
The method used which I do not understand is;
Suppose there is a rational zero , $r=\frac{m}{n}$ where $n$ is bigger than or equal to one, subbing in for $r$ and multiplying up by $n$ we get ,
$$2m^{3}-mn^2-6n^3,$$ if $p|n$ then from the equation $p|2m^3$. Since $p$ does not divide $m$, we have $p=2$, But then $4|2m^3$. So that $m$ and $n$ are both even; which is a contradiction. Thus $n=1$ and $m^{3}-m-6=0$ Hence $m|6$ and so we need to check the divisors of 6. Which shows it is irreducible over $\textbf{Q}$.
I understand the last bit once we have deduced $2m^{3}-m-6=0$ but it is the part before, please may someone explain this method.
Assume that a root is $\frac{m}{n}$ where $\gcd(m,n)=1$. Then as you show, we get $$2m^3 - mn^2 - 6n^3 = 0,\text{ so that }2m^3 = mn^2 + 6n^3 = n(mn+6n^2).$$ If $n\ne 1$, choose some prime $p\mid n$. Then $p$ divides the right-hand side above, so $p\mid 2m^3$. Since $\gcd(m,n) = 1$, we must have $p\mid 2$, so that $p = 2$. But then the right-hand side, $n(mn+6n^2)$, consists of two even factors, so is divisible by $4$. Thus $4\mid 2m^3$ and therefore $2\mid m$, a contradiction.
This implies $n=1$, so that (as you point out) we are reduced to looking for integral roots of $2m^3-m-6$.