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Question 8 : What is the largest positive integer such that \\frac{n^2+7n+12}{n^2-n-12}) is also positive integer?

- 6
- 8
- 16
- 12

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can be rewritten as

= 1 +

1 + = 1 +

n cannot be -3

has to be an integer

And this value has to be equal to 1, for n to be high

So,= 1

8 = n - 4

n = 12

The question is **"What is the largest positive integer such that \\frac{n^2+7n+12}{n^2-n-12}) is also positive integer? " **

**Choice D** is the correct answer.

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