Question

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

Answer 1

`color(blue)(x < -3)`

Explanation:

`x^3-3x^2-9x+27`
can be factored as
`color(white)("XXX")x^2(x-3)-9(x-3)`

`color(white)("XXX")(x^2-9)(x-3)`

`color(white)("XXX")(x+3)(x-3)(x-3)`

If `x < -3`
`color(white)("XXX")underbrace(underbrace(""(x-3))_("neg.")*underbrace(""(x-3))_("neg.")*underbrace(""(x+3))_("neg."))`
`color(white)("XXXXXXXX")< 0`

If `-3 < x < +3`
`color(white)("XXX")underbrace(underbrace(""(x-3))_("neg.")*underbrace(""(x-3))_("neg.")*underbrace(""(x+3))_("pos."))`
`color(white)("XXXXXXXX")> 0`

If `x > +3`
`color(white)("XXX")underbrace(underbrace(""(x-3))_("pos.")*underbrace(""(x-3))_("pos.")*underbrace(""(x+3))_("pos."))`
`color(white)("XXXXXXXX")> 0`

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

So the only case for
`color(white)("XXX")x^3-3x^2-9x+27 < 0`
is
`color(white)("XXX")x < -3`

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