The smallest integer $$n$$ such that $$n^3-11n^2+32n-280$$ is

Question

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

Accepted Answer

We can see that at n = 2, n 3 − 11 n 2 + 32 n − 28 = 0 n^3-11n^2+32n-28=0 n 3 − 11 n 2 + 32 n − 28 = 0 i.e. (n-2) is a factor of n 3 − 11 n 2 + 32 n − 28 n^3-11n^2+32n-28 n 3 − 11 n 2 + 32 n − 28 n 3 − 11 n 2 + 32 n − 28 n − 2 = n 2 − 9 n + 14 \dfrac{n^3-11n^2+32n-28}{n-2}=n^2-9n+14 n − 2 n 3 − 11 n

The smallest integer nnn such that n^3-11n^2+32n-28&gt;0 is

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