# How do you solve x^3-6x^2-4x+24>0 using a sign chart?

Dec 31, 2016

The answer is x in ] -2,+2 [ uu ] 6, +oo[

#### Explanation:

Let $f \left(x\right) = {x}^{3} - 6 {x}^{2} - 4 x + 24$

As a polynomial, the domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R}$

Then, by trial and error,

$f \left(2\right) = 8 - 24 - 8 + 24 = 0$

Therefore,

$\left(x - 2\right)$ is a factor

To find the other factors, we do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{3} - 6 {x}^{2} - 4 x + 24$$\textcolor{w h i t e}{a a a a}$∣$x - 2$

$\textcolor{w h i t e}{a a a a}$${x}^{3} - 2 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a a}$∣${x}^{2} - 4 x - 12$

$\textcolor{w h i t e}{a a a a}$$0 - 4 {x}^{2} - 4 x$

$\textcolor{w h i t e}{a a a a a a}$$- 4 {x}^{2} + 8 x$

$\textcolor{w h i t e}{a a a a a a a}$$- 0 - 12 x + 24$

$\textcolor{w h i t e}{a a a a a a a a a a}$$- 12 x + 24$

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

Therefore,

${x}^{2} - 4 x - 12 = \left(x + 2\right) \left(x - 6\right)$

and ${x}^{3} - 6 {x}^{2} - 4 x + 24 = \left(x - 2\right) \left(x + 2\right) \left(x - 6\right)$

Now, we can establish 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}$$- 2$$\textcolor{w h i t e}{a a a a a}$$2$$\textcolor{w h i t e}{a a a a a}$$6$$\textcolor{w h i t e}{a a a a}$$+ \infty$

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

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

$\textcolor{w h i t e}{a a a a}$$x - 6$$\textcolor{w h i t e}{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}$$+$

$\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}$$-$$\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}$$+$

Therefore,

$f \left(x\right) > 0$ when x in ] -2,+2 [ uu ] 6, +oo[