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

Given that, $n^{3} – 11n^{2} + 32n – 28 > 0 \quad \longrightarrow (1)$

For, the factorazation, let us assume

$n^{3} – 11n^{2} + 32n – 28 = 0 \quad \longrightarrow (2)$

Now, put the various of $n,$ and check

• $n = – 1 \Rightarrow (-1)^{3} – 11(-1)^{2} + 32(-1) – 28 = 0$

$\qquad \qquad \quad \; \Rightarrow -1 -11 -32 -28 = 0$

$\qquad \qquad \quad \; \Rightarrow \boxed{ -72 \neq 0}$

• $n=0 \Rightarrow (0)^{3} – 11 (0)^{2} + 32(0) – 28 = 0$

$\qquad \qquad \quad \; \Rightarrow \boxed{ -28 \neq 0}$

• $n=1 \Rightarrow (1)^{3} – 11 (1)^{2} + 32(1) – 28 = 0$

$\qquad \qquad \quad \; \Rightarrow 1 – 11 + 32 – 28 = 0$

$\qquad \qquad \quad \; \Rightarrow \boxed{ – 6 = 0}$

• $n=2 \Rightarrow (2)^{3} – 11 (2)^{2} + 32(2) – 28 = 0$

$\qquad \qquad \quad \; \Rightarrow 8 – 44 + 64 – 28 = 0$

$\qquad \qquad \quad \; \Rightarrow 72 – 72 = 0$

$\qquad \qquad \quad \; \Rightarrow \boxed{0 = 0}$

So, $(n-2)$ is the factor.

Now, we can divide the whole expression by $n-2.$

We can write, $(n-2) (n^{2} -9n + 14) = 0$

$\Rightarrow (n-2) ( n^{2} – 7n – 2n + 14) = 0$

$\Rightarrow (n -2) \left[n (n-7) -2 (n – 7) \right]$

$\Rightarrow (n – 2)(n – 2)(n – 7) = 0$

$\Rightarrow \boxed{n = 2, 2, 7}$

The $n = 7$,  makes the whole expression zero.

So, $n = 8$ is the smallest integer which make expression greater than zero.

Put the value of $n = 8,$ in the equation $(1),$ we get

$n^{3} – 11n^{2} + 32n – 28 > 0$

$\Rightarrow (8)^{3} – 11 (8)^{2} + 32 (8) – 28 > 0$

$\Rightarrow 512 – 704 + 256 – 28 > 0$

$\Rightarrow 768 – 732 > 0$

$\Rightarrow \boxed{36 > 0}$ (Satisfied)

$\therefore$ The value of smallest integer is $8.$

Correct Answer $:8$

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