For how many integers $n$ is $\frac{2 n^{3}-12 n^{2}-2 n+12}{n^{2}+5 n-6}$ equal to an integer?
Notice that $2(1)^{3}-12(1)^{2}-2(1)+12=0$ and $(1)^{2}+5(1)-6=0$, so 1 is a root of both the numerator and denominator.

This means the expression is not defined for $n=1$, but we can factor and divide both the numerator and denominator by $n-1$.

Doing this, we get

$$
\begin{aligned}
\frac{2 n^{3}-12 n^{2}-2 n+12}{n^{2}+5 n-6} & =\frac{(n-1)\left(2 n^{2}-10 n-12\right)}{(n-1)(n+6)} \\
& =\frac{2 n^{2}-10 n-12}{n+6}
\end{aligned}
$$

where the final equality holds as long as $n \neq 1$.

It is easily checked that -6 is not a root of the numerator, so no further simplification by cancellation is possible.

However, we can manipulate the expression to make it easier to understand for which $n$ the expression is equal to an integer.

We already know for integers $n \neq 1$ that the expression is equal to $\frac{2 n^{2}-10 n-12}{n+6}$.

This expression is not defined for $n=-6$, so we can further assume that $n \neq-6$ and rearrange to

$$
\begin{aligned}
\frac{2 n^{2}-10 n-12}{n+6} & =\frac{\left(2 n^{2}+12 n\right)-22 n-12}{n+6} \\
& =\frac{2 n(n+6)}{n+6}-\frac{22 n+12}{n+6} \\
& =2 n-\frac{(22 n+132)-120}{n+6} \\
& =2 n-\frac{22(n+6)}{n+6}+\frac{120}{n+6} \\
& =2 n-22+\frac{120}{n+6}
\end{aligned}
$$

The quantity $2 n-22$ is an integer for all integers $n$, so the original expression is an integer exactly when $\frac{120}{n+6}$ is an integer.

This means we need to count the integers $n$ for which $n+6$ is a divisor of 120 .

The divisors of 120 are

$$
\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 8, \pm 10, \pm 12, \pm 15, \pm 20, \pm 24, \pm 30, \pm 40, \pm 60, \pm 120
$$

of which there are 32 .

For each of these 32 divisors, we can solve for $n$. For example, if $n+6=-10$, then we have $n=-16$, so $n=-16$ is an integer for which the expression in the problem statement evaluates to an integer.

We also know that $n=1$ makes the expression undefined, so we need to make sure that solving for $n$ will never lead to $n=1$.

Notice that when $n=1, n+6=7$, and 7 is not a divisor of 120 .

Therefore, every divisor above gives an integer $n$ for which the expression is defined and equal to an integer, so the answer is 32 .

ANSWER: 32