# How do you solve x^3+4x^2-x>=4 using a sign chart?

Nov 17, 2016

The naswer is $x \in \left[- 4 , - 1\right] \cup \left[1 , + \infty\right]$

#### Explanation:

Let $f \left(x\right) = {x}^{3} + 4 {x}^{2} - x - 4$

Then, $f \left(1\right) = 1 + 4 - 1 - 4 = 0$

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

To find the other factors, let's do a long division

$\textcolor{w h i t e}{a a a a a}$${x}^{3} + 4 {x}^{2} - x - 4$$\textcolor{w h i t e}{a a a a a}$∣$x - 1$

$\textcolor{w h i t e}{a a a a a}$${x}^{3} - {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a a}$∣${x}^{2} + 5 x + 4$

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

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

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

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

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

Therefore, $f \left(x\right) = \left(x - 1\right) \left(x + 1\right) \left(x + 4\right) \ge 0$

Let's do the sign chart

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

$\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}$$x + 1$$\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 a}$$f \left(x\right)$$\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}$$+$

Therefore $f \left(x\right) \ge 0$ if $x \in \left[- 4 , - 1\right] \cup \left[1 , + \infty\right]$

graph{x^3+4x^2-x-4 [-20.28, 20.27, -10.13, 10.13]}