Consider $R=\frac{\mathbb{C}[X,Y,Z]}{(X^2-YZ)}.$ This is an integral domain since the ideal $(X^2-YZ)$ is prime, so we may talk about irreducibility in $R$. I'm interested in showing $x:=X+(X^2-YZ)$ is irreducible.
To this end, I suppose $x=fg$ for some $f,g$ in the quotient ring $R$ and aim to show $f$ or $g$ is a unit in $R$. Passing to $\mathbb{C}[X,Y,Z]$, this means we have $F,G,H\in \mathbb{C}[X,Y,Z]$ such that $$X=F\cdot G+H\cdot (X^2-YZ).$$ I want to now set $X=(YZ)^{1/2}$ to get that $$(YZ)^{1/2}=F((YZ)^{1/2},Y,Z)\cdot G((YZ)^{1/2},Y,Z)$$ and by a degree argument conclude that either $F$ or $G$ must be constant when evaluated at $X=(YZ)^{1/2}$, so must be of the form $c+(X^2-YZ)$ as desired. Thing is, I'm not sure why the only polynomials in $\mathbb{C}[X,Y,Z]$ that vanish when evaluated at $X=(YZ)^{1/2}$ are multiples of $X^2-YZ$.
Is this approach reparable? If not, how would you show $x$ is irreducible? Thanks.