Irreducibility of $x^n＋px＋p^2(n≧3)$ and newton polygon

It's sufficient for a polynomial to be irreducible in the larger field $\mathbb{Q}_p$ to imply that it's irreducible in $\mathbb{Q}$. However in this case, it is reducible in $\mathbb{Q}_p$ so you can't draw that conclusion. A similar, simpler example is we know $x^2+1$ is irreducible in $\mathbb{Q}$, but of course it's reducible in $\mathbb{Q}_5$ because we have two elements that square to $-1$.

That being said, here's a way to see the specific root and where to start lifting from when $n\ge 3$. The reason is from the Newton polygon we see there is a root with p-adic absolute value $p^{-1}$, so we can guess a little with the general Hensel lemma to see that $x=-p$ works and satisfies the following condition to be lifted,

$$|f(x)|_p<|f'(x)|_p^2$$ $$|x^n+px+p^2|_p<|nx^{n-1}+p|_p^2$$ Now letting $x=-p$ be our guess, $$|(-p)^n|_p<|n(-p)^{n-1}-p|_p^2$$ The ultrametric inequality forces the equality $|n(-p)^{n-1}-p|_p=|p|_p$ $$p^{-n}<p^{-2}$$ This means this works for the case when $n>2$ as desired.