# How do you solve (2y^2+3y-20)/(y^3-y^2)>0?

Jan 11, 2017

The answer is y in [-4, 0[ uu] 0,1 [uu [5/2, +oo[

#### Explanation:

Factorise the numerator and denominator

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

${y}^{3} - {y}^{2} = {y}^{2} \left(y - 1\right)$

So,

$\frac{2 {y}^{2} + 3 y - 20}{{y}^{3} - {y}^{2}} = \frac{\left(2 y - 5\right) \left(y + 4\right)}{{y}^{2} \left(y - 1\right)}$

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

The domain of $y$ is ${D}_{y} = \mathbb{R} - \left\{0 , 1\right\}$

${y}^{2} > 0 \forall y \in {D}_{y}$

Let's do a sign chart

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 4$$\textcolor{w h i t e}{a a a a a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$1$$\textcolor{w h i t e}{a a a a a}$$\frac{5}{2}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$y + 4$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$color(red)(∥)$\textcolor{w h i t e}{}$$+$color(red)(∥)$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$y - 1$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$color(red)(∥)$\textcolor{w h i t e}{}$$-$color(red)(∥)$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$2 y - 5$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$color(red)(∥)$\textcolor{w h i t e}{}$$-$color(red)(∥)$\textcolor{w h i t e}{a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(y\right)$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$color(red)(∥)$\textcolor{w h i t e}{}$$+$color(red)(∥)$\textcolor{w h i t e}{a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(y\right) > 0$, when y in [-4, 0[ uu] 0,1 [uu [5/2, +oo[