How to prove a vanishing set is a subset of $\mathbb{C}P^1 \times \mathbb{C}P^1$

I've been dabbling in algebraic geometry lately and studying some of the basic concepts. I understand that, if I have a homogeneous polynomial

$$ F \in \mathbb{C}[X_0,X_1,\dots, X_{n}] $$

then the vanishing set $V(F)$ is a subset of $\mathbb{C}P^n$. However, I've also found other kind of polynomials such as this one:

$$ F = X_0^2 Y_0^3 + X_1^2 Y_1^3 + X_0 X_1 (Y_0^3 - Y_1^3) \in \mathbb{C}[X_0,X_1,Y_0,Y_1]. $$

This polynomial is a degree $5$ homogeneous polynomial in four variables, and so should be a hypersurface in $\mathbb{C}P^3$. In addition, it is a bihomogeneous polynomial of degree $2$ in $(X_0,X_1)$ and degree $3$ in $(Y_0,Y_1)$. I've been told that vanishing sets of polynomials of this form are also subsets of $\mathbb{C}P^1 \times \mathbb{C}P^1$ of dimension $1$. However, I fail to see how that is the case. I can't express $F$ as a product of a polynomial in $\mathbb{C}[X_0,X_1]$ and another one in $\mathbb{C}[Y_0,Y_1]$, and I fail to see another way to prove this.

Thanks for any help in advance.

There are two different vanishing sets. One is a subset of $\mathbb{CP}^3$, and the other is a subset of $\mathbb{CP}^1\times\mathbb{CP}^1$:

$$\{[X_0, X_1, Y_0, Y_1] \in \mathbb{CP}^3 \mid F(X_0, X_1, Y_0, Y_1) = 0\} \subset \mathbb{CP}^3$$

$$\{([X_0, X_1], [Y_0, Y_1]) \in \mathbb{CP}^1\times\mathbb{CP}^1 \mid F(X_0, X_1, Y_0, Y_1) = 0\} \subset \mathbb{CP}^1\times\mathbb{CP}^1.$$

In general, a homogeneous polynomial has a well-defined vanishing set in projective space, and a multi-homogeneous polynomial has a well-defined vanishing set in a product of projective spaces. Every multi-homogeneous polynomial is homogeneous, so you always get two vanishing sets.