What would be the Weierstrass factorisation of $f(x)=p-x+px^2$? To find the Weierstrass factorisation of the polynomial $f(x)=p-x+px^2$, we solve for the roots $ \alpha=\frac{1 \pm \sqrt{1-4p^2}}{2p}$. Now the square roots converges in $p$-adic ring integer $\mathbb{Z}_p$ because the binomial series $(1-4p^2)^{1/2}$ converges in $\mathbb{Z}_p$. 
Let $\alpha$ and $\beta$ be the two roots. The roots are related by $\alpha \beta=1$. 
My questions:


*

*Is any of these roots integral?

*What would be the Weierstrass factorisation?


I hope it is not simply $f(x)=(x-\alpha)(x-\beta)$. It should be $f(x)=p(x) g(x)$ for some monic polynomial $p(x)$ and $g(x)$ being unit power series over $\mathbb{Z}_p$.  
If I take $p(x)=x-\alpha$ to be the monic polynomial, then the only option left for $g(x)$ to be unit power series in $\mathbb{Z}_p$ is $g(x)=x-p\beta$ as $p \beta$ is unit in $\mathbb{Z}_p$.  
Please answer the above two questions. 
 A: Best look at the Newton Polygon, which is formed by associating to each monomial $ax^m$ the point $(n,v(a))$ in $\Bbb R^2$ and erecting on each the vertical upward ray. Then take the convex hull of the union of these rays. Mostly, you’re interested in the lower boundary of this figure.
Now, each segment of the boundary will between between two points $(n_i,v(a_i))$ and $(n_{i'},v(a_{i'})$ with $n_i<n_{i'}$; it will have the width $w_{ii'}=n_{i'}-n_i$ and the negative slope$$s_{ii'}=\frac{v(a_{i'})-v(a_i)}{n_i-n_{i'}}\,. $$
Then the roots of the polynomial are: $w_{ii'}$ of them have valuation $s_{ii'}$. You can prove all of this fairly easily, using the fact that if a sum of finitely many terms is zero, then two of the terms must have the same, minimum, valuation.
Now, in the case of your quadratic polynomial, the vertices of the Polygon are $(0,1)$, $(1,0)$, and $(2,1)$. The segments are both of width $1$, slopes $-1$ and $1$. Thus one root has valuation $1$, the other valuation $-1$. Indeed, if you divide your polynomial by $p$, it becomes $x^2-x/p+1$. So each root is the reciprocal of the other. And you could have seen that just at the outset, without the Polygon. But if you want to know about roots of $p$-adic polynomials, you must understand Newton Polygon.
Anyway, if $\alpha$ is the integral root, and $1/\alpha$ the other, your Weierstrass factorization is $(x-\alpha)(px-p/\alpha)$.
