Largest integer for algebraic fraction

Simplify the given algebraic fraction: $\frac{n^2+7n+12}{n^2-n-12} = \frac{(n+3)(n+4)}{(n-4)(n+3)} = \frac{n+4}{n-4} \quad (\text{for } n \neq -3)$

Rewrite as: $\frac{n+4}{n-4} = 1 + \frac{8}{n-4}$

For this to be a positive integer, $(n-4)$ must be a divisor of 8. To maximize $n$, set $n - 4$ to the largest positive divisor of 8, which is 8: $n - 4 = 8 \implies n = 12$

For $n = 12$, $\frac{12+4}{12-4} = \frac{16}{8} = 2$, which is a positive integer.