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For how many integer values of \(n\) is \(\frac{n^2 + 6n + 24}{n + 3}\) an integer?
A. 1
B. 2
C. 4
D. 6
E. 8
A. 1
B. 2
C. 4
D. 6
E. 8
23 Jan 2024, 02:55
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Official Solution:
For how many integer values of \(n\) is \(\frac{n^2 + 6n + 24}{n + 3}\) an integer?
A. 1
B. 2
C. 4
D. 6
E. 8
Start with completing the square in the numerator:
Split the fraction:
Since \(n + 3\) is an integer, the whole expression will be an integer whenever \(\frac{15}{n + 3}\) is an integer.
For \(\frac{15}{n + 3}\), the denominator, \(n+3\), must be a divisor of 15. Since \(15 = 3*5\), it has \((1 + 1)(1 + 1) = 4\) positive divisors and 4 negative ones: 1, 3, 5, 15, -1, -3, -5, and -15. Therefore, \(n\) can take a total of 8 values: -2, 0, 2, 12, -4, -6, -8, and -18, respectively.
Answer: E
For how many integer values of \(n\) is \(\frac{n^2 + 6n + 24}{n + 3}\) an integer?
A. 1
B. 2
C. 4
D. 6
E. 8
Start with completing the square in the numerator:
\(\frac{n^2 + 6n + 24}{n + 3} = \frac{n^2 + 6n + 9 + 15}{n + 3} = \frac{(n + 3)^2 + 15}{n + 3}\).
Split the fraction:
\(\frac{(n + 3)^2 + 15}{n + 3} = \frac{(n + 3)^2}{n + 3} + \frac{15}{n + 3} = n + 3 + \frac{15}{n + 3}\).
Since \(n + 3\) is an integer, the whole expression will be an integer whenever \(\frac{15}{n + 3}\) is an integer.
For \(\frac{15}{n + 3}\), the denominator, \(n+3\), must be a divisor of 15. Since \(15 = 3*5\), it has \((1 + 1)(1 + 1) = 4\) positive divisors and 4 negative ones: 1, 3, 5, 15, -1, -3, -5, and -15. Therefore, \(n\) can take a total of 8 values: -2, 0, 2, 12, -4, -6, -8, and -18, respectively.
Answer: E
18 Jan 2025, 01:13
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I like the solution - it’s helpful.
