# How do you solve x^3-5x^2<x-5 using a sign chart?

Feb 13, 2017

The answer is x in ]-oo, -1 [uu]1,5[

#### Explanation:

Let's rewrite the inequality

${x}^{3} - 5 {x}^{2} < x - 5$

${x}^{3} - 5 {x}^{2} - x + 5 < 0$

Let $f \left(x\right) = {x}^{3} - 5 {x}^{2} - x + 5$

$f \left(1\right) = 1 - 5 - 1 + 5 = 0$

so,

$\left(x - 1\right)$ is a factor of $f \left(x\right)$

$f \left(- 1\right) = - 1 - 5 + 1 + 5 = 0$

so,

$\left(x + 1\right)$ is a factor of $f \left(x\right)$

So,

$\left(x + 1\right) \left(x - 1\right) = {x}^{2} - 1$ is a factor of $f \left(x\right)$

We perform a long division to find the 3rd factor

$\textcolor{w h i t e}{a a a a}$${x}^{3} - 5 {x}^{2} - x + 5$$\textcolor{w h i t e}{a a a a}$$|$${x}^{2} - 1$

$\textcolor{w h i t e}{a a a a}$${x}^{3}$$\textcolor{w h i t e}{a a a a a a}$$- x$$\textcolor{w h i t e}{a a a a a a a a}$$|$$x - 5$

$\textcolor{w h i t e}{a a a a}$$0 - 5 {x}^{2}$$\textcolor{w h i t e}{a a a}$$0 + 5$

$\textcolor{w h i t e}{a a a a a a}$$- 5 {x}^{2}$$\textcolor{w h i t e}{a a a a a}$$+ 5$

$\textcolor{w h i t e}{a a a a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a}$$+ 0$

Therefore,

$f \left(x\right) = \left(x + 1\right) \left(x - 1\right) \left(x - 5\right)$

Let's build the sign chart

$\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}$$1$$\textcolor{w h i t e}{a a a a a}$$5$$\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}$$x - 1$$\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}$$x - 5$$\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 ]-oo, -1 [uu]1,5[