Irreducibility of $x^n＋px＋p^2(n≧3)$ and newton polygon It is well known that

If $f$ is irreducible polynomial over $\Bbb Q_p$, then there is only
one slope, i.e. the newton polygon of $f$ consists of a single
segment.

Now, newton polygon of $x^n＋px＋p^2(n≧3)$ consists of two segment, so from above, it turns out that $f$ is reducible.
But in fact, $f$ is irreducible.
Where am I missing ?
Thank you for your kind help.
 A: 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.
