Question

The smallest integer n such that n3 - 11n2 + 32n - 28 > 0 is

Answer 1

Correct Answer: 8

To find the smallest integer \( n \) such that \( n^3 - 11n^2 + 32n - 28 > 0 \), we follow these steps:

1. Evaluate the polynomial for initial values close to the suspected range.
2. Check for the smallest integer satisfying the inequality.

First, substitute incremental integer values starting from \( n = 8 \):

n	f(n) = \( n^3 - 11n^2 + 32n - 28 \)	f(n) > 0?
8	\( 8^3 - 11 \times 8^2 + 32 \times 8 - 28 = 512 - 704 + 256 - 28 = 36 \)	Yes
7	\( 7^3 - 11 \times 7^2 + 32 \times 7 - 28 = 343 - 539 + 224 - 28 = 0 \)	No, equals 0

• The polynomial evaluates to 36 (which is > 0) when \( n = 8 \). At \( n = 7 \), it evaluates to exactly 0, which is not greater than 0.
• Thus, the smallest integer \( n \) such that \( f(n) > 0 \) is \( n = 8 \).

The computed value \( n = 8 \) falls within the provided range: \([8, 8]\).