Prove that if the coefficient of $ax^3+bx+c$ are odd then it is irreducible of $\mathbb{Q}$

Question

Let $a,b,c$ be odd integers. Prove that $p(x)=ax^3+bx+c$ has no rational root.

Answer 1

Hint: Suppose to the contrary that there is a rational root $\dfrac{p}{q}$. We can assume that $p$ and $q$ are relatively prime. So at least one of them is odd. Substituting, we get $ap^2+bpq+cq^2=0$.

This is impossible. If one of $p$ or $q$ is odd and the other even, then $ap^2+bpq+cq^2$ is odd. This is also the case if both $p$ and $q$ are both odd. So in neither case can $ap^2+bpq+cq^2$ be equal to $0$.

Answer 2

Assume that the reduced fraction $p/q$ is a solution. Then $$ap^3+bpq^2+cq^3=0.$$

Then $p$ is a divisor of $c$ and $q$ is a divisor of $a$. Then both are odd. But then the above is a sum of $3$ odd numbers.