# How do you solve (12n^3+16n^2-3n-4)/(8n^3+12n^2+10n+15)>0?

##### 1 Answer
Mar 2, 2018

The solution is $n \in \left(- \infty , - \frac{3}{2}\right) \cup \left(- \frac{4}{3} , - \frac{1}{2}\right) \cup \left(\frac{1}{2} , + \infty\right)$

#### Explanation:

Start by factorising the numerator and the denominator

$12 {n}^{3} + 16 {n}^{2} - 3 n - 4 = 12 {n}^{3} - 3 n + 16 {n}^{2} - 4$

$= 3 n \left(4 {n}^{2} - 1\right) + 4 \left(4 {n}^{2} - 1\right)$

$= \left(4 {n}^{2} - 1\right) \left(3 n + 4\right)$

$= \left(2 n + 1\right) \left(2 n - 1\right) \left(3 n + 4\right)$

$8 {n}^{3} + 12 {n}^{2} + 10 n + 15 = 8 {n}^{3} + 10 n + 12 {n}^{2} + 15$

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

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

Finally,
$\frac{12 {n}^{3} + 16 {n}^{2} - 3 n - 4}{8 {n}^{3} + 12 {n}^{2} + 10 n + 15} = \frac{\left(2 n + 1\right) \left(2 n - 1\right) \left(3 n + 4\right)}{\left(4 {n}^{2} + 5\right) \left(2 n + 3\right)}$

Let $f \left(n\right) = \frac{\left(2 n + 1\right) \left(2 n - 1\right) \left(3 n + 4\right)}{\left(4 {n}^{2} + 5\right) \left(2 n + 3\right)}$

The term $\left(4 {n}^{2} + 5\right) > 0$

We can construct the sign chart

$\textcolor{w h i t e}{a a a a}$$n$$\textcolor{w h i t e}{a a a}$$- \infty$$\textcolor{w h i t e}{a a a}$$- \frac{3}{2}$$\textcolor{w h i t e}{a a a}$$- \frac{4}{3}$$\textcolor{w h i t e}{a a a a}$$- \frac{1}{2}$$\textcolor{w h i t e}{a a a a}$$\frac{1}{2}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$2 n - 3$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$3 n + 4$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$2 n + 1$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$2 n - 1$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(n\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(n\right) > 0$ when $n \in \left(- \infty , - \frac{3}{2}\right) \cup \left(- \frac{4}{3} , - \frac{1}{2}\right) \cup \left(\frac{1}{2} , + \infty\right)$

graph{(12x^3+16x^2-3x-4)/(8x^3+12x^2+10x+15) [-5.944, 3.92, -2.265, 2.665]}