Is this a valid proof that $\sqrt{3}$ is irrational?

Question

Consider the following polynomial $P(x) = x^2 - 3$. By the rational root theorem, the only possible rational roots are $\pm{3}$ or $\pm{1}$. Substituting these into $P(x)$ gives $6$ and $-2$, respectively. Thus $P(x)$ has no rational roots. Thus $\sqrt{3}$ is not rational.