The smallest integer $n$ such that $n^3 - 11n^2 + 32n - 28 > 0$ is

We need to find the smallest integer $n$ that satisfies the inequality $n^3 - 11n^2 + 32n - 28 > 0$.

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The key insight is to factor the polynomial $n^3 - 11n^2 + 32n - 28$.

Let's check if $n = 2$ is a root:

$2^3 - 11(2^2) + 32(2) - 28 = 8 - 44 + 64 - 28 = 0$

Since $n = 2$ makes the polynomial equal to zero, $(n-2)$ is a factor.

Using polynomial division, we find that:

$n^3 - 11n^2 + 32n - 28 = (n-2)^2(n-7)$

Notice that $n = 2$ is actually a repeated root (appears twice), which is why we get $(n-2)^2$ in the factorization.

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Our inequality becomes:

$(n-2)^2(n-7) > 0$

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To solve this inequality, we need to understand when the product $(n-2)^2(n-7)$ is positive.

Key insight: $(n-2)^2$ is always non-negative for any real number $n$, and equals zero only when $n = 2$.

Let's analyze the sign of $(n-7)$:

When $n < 7$: $(n-7) < 0$ (negative)

When $n = 7$: $(n-7) = 0$ (zero)

When $n > 7$: $(n-7) > 0$ (positive)

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For the product $(n-2)^2(n-7) > 0$ to be positive:

When $n < 2$: $(n-2)^2$ is positive and $(n-7)$ is negative, so the product is negative

When $n = 2$: $(n-2)^2$ is zero and $(n-7)$ is negative, so the product is zero

When $2 < n < 7$: $(n-2)^2$ is positive and $(n-7)$ is negative, so the product is negative

When $n = 7$: $(n-2)^2$ is positive and $(n-7)$ is zero, so the product is zero

When $n > 7$: $(n-2)^2$ is positive and $(n-7)$ is positive, so the product is positive

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The inequality $(n-2)^2(n-7) > 0$ is satisfied when $n > 7$.

The smallest integer greater than 7 is 8.

Therefore, the smallest integer $n$ is 8.