Resolving nodes of a quintic CY 3-fold Let's consider the following quintic 3-fold $X$:
\begin{equation}
\{(x_i) \in \mathbb{P}^4 \ | \ x_1f(x)-x_2g(x)=0\}
\end{equation}
for generic homogeneous polynomials $f(x),g(x)$ of degree four. It is easy to se that $X$ has $16$ nodes at $\{x_1=x_2=f(x)=g(x)=0\}$. I now want a crepant resolution of $X$. There are four ways to blow-up the singularities, namely blow-up one of the four divisors:
\begin{equation}
\{x_1=x_2=0\},\ \{x_1=g(x)=0\},\ \{f(x)=x_2=0\},\ \{f(x)=g(x)=0\}.
\end{equation}
Is it true that any one of them gives a K\"{a}hler crepant resolution of $X$ and that the resulting Hodge numbers do not depend on the choice of resolutions?
 A: I assume that you mean $x_1=x_2=0$ in your description of the divisors. I believe that you only really get two different resolutions, but yes, they are both projective varieties. Hodge numbers of birational complex Calabi-Yau varieties are always the same.
You can think of this as a determinantal variety for the matrix $A$ of the form 
$$
\Big(\begin{array}{cc}{\rm deg~1}& {\rm deg~1}\\
{\rm deg~4}& {\rm deg~4}
\end{array}\Big)
$$
in the variables $x$. The resolutions correspond to considering degenerate matrices together with either left or right kernel vectors.
This is also very much a toric example, in the sense that both resolutions are complete intersections of two hypersurfaces that correspond to "double-mirror" nef partitions, see for example the paper of my student Zhan Li "On the birationality of complete intersections associated to nef-partitions",  arXiv:1310.2310.
Specifically, if I got my calculations right, one resolution is a complete intersection
of degrees $(1,1)$ and $(1,4)$ in $\mathbb P^1\times \mathbb P^4$, and the other is a
complete intersection in the $\mathbb P^1$ bundle $\mathbb P(\mathcal O(-1)+\mathcal O(-4)$) over $\mathbb P^4$ of the half-anticanonical divisor that corresponds to the tautological line bundle $\mathcal O(1)$ of this fibration. The singular variety in question is the projection to the $\mathbb P^4$ in both cases.
