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

Jul 7, 2016

$\textcolor{b l u e}{x < - 3}$

#### Explanation:

${x}^{3} - 3 {x}^{2} - 9 x + 27$
can be factored as
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} \left(x - 3\right) - 9 \left(x - 3\right)$

$\textcolor{w h i t e}{\text{XXX}} \left({x}^{2} - 9\right) \left(x - 3\right)$

$\textcolor{w h i t e}{\text{XXX}} \left(x + 3\right) \left(x - 3\right) \left(x - 3\right)$

If $x < - 3$
color(white)("XXX")underbrace(underbrace(""(x-3))_("neg.")*underbrace(""(x-3))_("neg.")*underbrace(""(x+3))_("neg."))
$\textcolor{w h i t e}{\text{XXXXXXXX}} < 0$

If $- 3 < x < + 3$
color(white)("XXX")underbrace(underbrace(""(x-3))_("neg.")*underbrace(""(x-3))_("neg.")*underbrace(""(x+3))_("pos."))
$\textcolor{w h i t e}{\text{XXXXXXXX}} > 0$

If $x > + 3$
color(white)("XXX")underbrace(underbrace(""(x-3))_("pos.")*underbrace(""(x-3))_("pos.")*underbrace(""(x+3))_("pos."))
$\textcolor{w h i t e}{\text{XXXXXXXX}} > 0$

Note if $x = - 3$ or $x = + 3$ the result $= 0$

So the only case for
$\textcolor{w h i t e}{\text{XXX}} {x}^{3} - 3 {x}^{2} - 9 x + 27 < 0$
is
$\textcolor{w h i t e}{\text{XXX}} x < - 3$

graph{x^3-3x^2-9x+27 [-65, 66.65, -31.1, 34.75]}