# How do you solve x^3-3x^2-9x+27<0 using a sign chart?

Nov 12, 2016

The answer is x in ] -oo,-3 [

#### Explanation:

Let $f \left(x\right) = {x}^{3} - 3 {x}^{2} - 9 x + 27$ and $f \left(x\right) < 0$

Then, $f \left(- 3\right) = - 27 - 27 + 27 + 27 = 0$

So, $\left(x + 3\right)$ is a factor of $f \left(x\right)$

Let's do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{3} - 3 {x}^{2} - 9 x + 27$$\textcolor{w h i t e}{a a a a}$∣$x + 3$
$\textcolor{w h i t e}{a a a a}$${x}^{3} + 3 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a a}$∣${x}^{2} - 6 x + 9$
$\textcolor{w h i t e}{a a a a a}$$0 - 6 {x}^{2} - 9 x$
$\textcolor{w h i t e}{a a a a a a a}$$- 6 {x}^{2} - 18 x$
$\textcolor{w h i t e}{a a a a a a a a a a a}$$0 + 9 x + 27$
$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$+ 9 x + 27$

${x}^{2} - 6 x + 9 = {\left(x - 3\right)}^{2}$

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

let's do the sign chart

$\textcolor{w h i t e}{a 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 a}$$- 3$$\textcolor{w h i t e}{a a a a a}$$3$$\textcolor{w h i t e}{a a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a a}$$x + 3$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$
$\textcolor{w h i t e}{a a a a a}$${\left(x - 3\right)}^{2}$$\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 a}$$+$
$\textcolor{w h i t e}{a a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$

$f \left(x\right) < 0$ when x in ] -oo,-3 [

graph{x^3-3x^2-9x+27 [-76.4, 55.3, -32.1, 33.74]}