# How do you solve (n^3+3n^2-4n-12)/(n^3-5n^2+4n-20)<=0?

Jan 9, 2017

$n \in \left[- 3 , - 2\right] \cup \left[2 , 5\right)$

#### Explanation:

${n}^{3} + 3 {n}^{2} - 4 n - 12 = \left({n}^{3} + 3 {n}^{2}\right) - \left(4 n + 12\right)$

$\textcolor{w h i t e}{{n}^{3} + 3 {n}^{2} - 4 n - 12} = {n}^{2} \left(n + 3\right) - 4 \left(n + 3\right)$

$\textcolor{w h i t e}{{n}^{3} + 3 {n}^{2} - 4 n - 12} = \left({n}^{2} - 4\right) \left(n + 3\right)$

$\textcolor{w h i t e}{{n}^{3} + 3 {n}^{2} - 4 n - 12} = \left({n}^{2} - {2}^{2}\right) \left(n + 3\right)$

$\textcolor{w h i t e}{{n}^{3} + 3 {n}^{2} - 4 n - 12} = \left(n - 2\right) \left(n + 2\right) \left(n + 3\right)$

$\textcolor{w h i t e}{}$

${n}^{3} - 5 {n}^{2} + 4 n - 20 = \left({n}^{3} - 5 {n}^{2}\right) + \left(4 n - 20\right)$

$\textcolor{w h i t e}{{n}^{3} - 5 {n}^{2} + 4 n - 20} = {n}^{2} \left(n - 5\right) + 4 \left(n - 5\right)$

$\textcolor{w h i t e}{{n}^{3} - 5 {n}^{2} + 4 n - 20} = \left({n}^{2} + 4\right) \left(n - 5\right)$

$\textcolor{w h i t e}{}$
So the numerator and denominator have no common factors and any linear factors occur precisely once.

The zeros of one or other are at $- 3 , - 2 , 2 , 5$, with $n = 5$ being a vertical asymptote.

So the quotient changes sign at each of these points.

Note that for large positive values of $n$, the signs of both the numerator and denominator are both positive, so their quotient is positive too.

Hence the sign of the quotient in each of the following intervals is as follows:

$\left(5 , \infty\right) : +$

$\left(2 , 5\right) : -$

$\left(- 2 , 2\right) : +$

$\left(- 3 , - 2\right) : -$

$\left(- \infty , - 3\right) : +$

Hence the inequality is satisfied for:

$n \in \left[- 3 , - 2\right] \cup \left[2 , 5\right)$

graph{ (x^3+3x^2-4x-12)/(x^3-5x^2+4x-20) [-8.085, 11.915, -4.76, 5.24]}