Question

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

Answer 1

Answer:

8

Step-by-step explanation:

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

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

$(n-2)^2\geq 0$

∴$(n-2)^2> 0$ $(n\neq 2)$

$n-7>0$

∴$n>7$

The smallest integer n is 8.

∴ $n=8$

Question solved!

Answer 2

Step-by-step explanation:

I think the answer will be 0