# Double zeros of a polynomial in two variables

Let $F\in\mathbb{C}[X,Y]$ be an irreducible polynomial and $n\in \mathbb{N}$, $n\ge1$ ,$p_i\in\mathbb{C}[X]$ for $0\le i\le n$, such that

$$F(X,Y)=\sum\limits_{i=0}^{n}p_i(X)Y^{n-i}$$

Let $x\in\mathbb{C}$, such that $p_0(x)\ne0$ (so $F(x,Y)$ has n zeros) and one of this $n$ zeros is a single one (so there is a $y\in\mathbb{C}$, such that $F(x,y)=0$ and $\frac{\partial F}{\partial Y}(x,y)\ne0$). Is it possible, that the polynomial $F(x,Y)$ has a multiple (double or more) zero?

[I noticed that i swapped $x$ and $y$, but i don't think this is a big issue] $F(x,y)=x^3-yx^2-y^2x+1$ is irreducible: $$(a(x)y+b(x))(c(x)y+d(x))=a(x)c(x)y^2+y(a(x)d(x)+b(x)c(x))+b(x)d(x)$$ gives $a(x)c(x)=x$, hence wlog $a(x)=1$, $c(x)=x$, so $d(x)+xb(x)=x^2$ and $b(x)d(x)=x^3+1=(x+1)(x^2-x+1)$ If $b(x)=x+1$, then $d(x)=x^2-x^2-x=-x$ and it doesn't work. If $b(x)=x^2-x+1$, then $d(x)=x^2-x^3+x^2-x=-x^3+2x^2-x$ and it doesn't work.
If $y=1$, $F(x,1)=x^3-x^2-x+1=(x+1)(x-1)^2$, so it has a single root and a double root.