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Actually, there is a rather straightforward way to see that this singularity is factorial:

Let $X=\mathrm{Spec}\ k[x_1,x_2,x_3,x_4,x_5]/(x_1x_2+x_3x_4+x_5^2)$ and let $D=(x_2=0)\subseteq X$ be a principal divisor. Notice that then $D= \mathrm{Spec}\ k[x_1,x_3,x_4,x_5]/(x_3x_4+x_5^2)$, so it is a cone over a quadric cone. In particular, it is irreducible and reduced and hence one has an exact sequence: $$\mathbb Z\cdot D \to \mathrm{Cl}(X) \to \mathrm{Cl}(X\setminus D)\to 0.$$

Since $D$ is principal, it is actually zero in $\mathrm{Cl}(X)$, so this says that $$\mathrm{Cl}(X) \simeq \mathrm{Cl}(X\setminus D)$$

On the other hand, it is easy to see that $$X\setminus D=\mathrm{Spec}\ k[x_1,x_2,x_2^{-1},x_3,x_4,x_5]/(x_1x_2+x_3x_4+x_5^2)\simeq \mathrm{Spec}\ k[x_2,x_2^{-1},x_3,x_4,x_5].$$ Clearly, $k[x_2,x_2^{-1},x_3,x_4,x_5]$ is a UFD, so $\mathrm{Cl}(X) \simeq \mathrm{Cl}(X\setminus D)=0$. Therefore the singularity in question is indeed factorial. Then, as Karl has already pointed out, it follows for instance from the argument here that it does not admit a small resolution of singularities .and since it is terminal then also all of its resolutions are non-crepant (I guess perhaps the correct word would be "discrepant").

In fact, the above argument shows the same statement for any singularity defined by a non-degenerate quadric equation in at least 5 variables.

It may be worth noting that the argument fails for fewer number of variables because $D$ is either not irreducible (in 4 variables) or reduced (in 3 variables). In 2 variables X itself is not irreducible.

Finally, let me add that I find it odd to call this a Calabi-Yau 4fold. I know that this terminology is used by others (especially physicists) as well, but I think it is rather misleading. This is at best a log Calabi-Yau.

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Actually, there is a rather straightforward way to see that this singularity is factorial:

Let $X=\mathrm{Spec}\ k[x_1,x_2,x_3,x_4,x_5]/(x_1x_2+x_3x_4+x_5^2)$ and let $D=(x_2=0)\subseteq X$ be a principal divisor. Notice that then $D= \mathrm{Spec}\ k[x_1,x_3,x_4,x_5]/(x_3x_4+x_5^2)$, so it is a cone over a quadric cone. In particular, it is irreducible and reduced and hence one has an exact sequence: $$\mathbb Z\cdot D \to \mathrm{Cl}(X) \to \mathrm{Cl}(X\setminus D)\to 0.$$

Since $D$ is principal, it is actually zero in $\mathrm{Cl}(X)$, so this says that $$\mathrm{Cl}(X) \simeq \mathrm{Cl}(X\setminus D)$$

On the other hand, it is easy to see that $$X\setminus D=\mathrm{Spec}\ k[x_1,x_2,x_2^{-1},x_3,x_4,x_5]/(x_1x_2+x_3x_4+x_5^2)\simeq \mathrm{Spec}\ k[x_2,x_2^{-1},x_3,x_4,x_5].$$ Clearly, $k[x_2,x_2^{-1},x_3,x_4,x_5]$ is a UFD, so $\mathrm{Cl}(X) \simeq \mathrm{Cl}(X\setminus D)=0$. Therefore the singularity in question is indeed factorial. Then, as Karl has already pointed out, it follows for instance from the argument here that it does not admit a resolution of singularities.

In fact, the above argument shows the same statement for any singularity defined by a non-degenerate quadric equation in at least 5 variables.

It may be worth noting that the argument fails for fewer number of variables because $D$ is either not irreducible (in 4 variables) or reduced (in 3 variables). In 2 variables X itself is not irreducible.

Finally, let me add that I find it odd to call this a Calabi-Yau 4fold. I know that this terminology is used by others (especially physicists) as well, but I think it is rather misleading. This is at best a log Calabi-Yau.