# How do you solve and write the following in interval notation: x^3-3x^2-4x<0?

Aug 25, 2017

The solution is $x \in \left(- \infty , - 1\right) \cup \left(0 , 4\right)$

#### Explanation:

We start by factorising the inequality

${x}^{3} - 3 {x}^{2} - 4 x < 0$

$x \left({x}^{2} - 3 x + 4\right) < 0$

$x \left(x + 1\right) \left(x - 4\right) < 0$

Let $f \left(x\right) = x \left(x + 1\right) \left(x - 4\right)$

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a a}$$4$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a 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}$$+$

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a 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}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 4$$\textcolor{w h i t e}{a a 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}$$+$

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

Therefore,

$f \left(x\right) < 0$ when $x \in \left(- \infty , - 1\right) \cup \left(0 , 4\right)$

graph{x^3-3x^2-4x [-20, 20.55, -14.88, 5.4]}