Question

What is the largest positive integer n such that \(\frac{n^2+7n+12}{n^2-n-12}\)  is also a positive integer?

• A. 8
• B. 12
• C. 16
• D. 6

Answer 1

The Correct Option is B

We are given the rational expression: 

\[ \frac{n^2 + 7n + 12}{n^2 - n - 12} \]

Step 1: Factorize numerator and denominator

Factor the numerator: \[ n^2 + 7n + 12 = (n+3)(n+4) \]

Factor the denominator: \[ n^2 - n - 12 = (n+3)(n-4) \]

Step 2: Cancel common terms

Now simplify the expression: \[ \frac{(n+3)(n+4)}{(n+3)(n-4)} = \frac{n+4}{n-4} \]

Step 3: Express in the form of integer + fraction

We rewrite: \[ \frac{n+4}{n-4} = \frac{(n-4) + 8}{n-4} = 1 + \frac{8}{n-4} \]

Step 4: Ensure expression is a positive integer

For the entire expression to be a positive integer, the fractional part \[ \frac{8}{n-4} \] must be an integer. This implies: \[ n - 4 \mid 8 \]

Step 5: Find values that divide 8

The positive integer divisors of 8 are: 1, 2, 4, and 8.

Thus, possible values of \( n \) are: \[ n = 5, 6, 8, 12 \]

Step 6: Maximize the value of \( n \)

The maximum value from this list is: \[ \boxed{12} \]

Answer:

The maximum value of \( n \) such that the given expression is a positive integer is 12.