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

1 Answer
Alan P.
Jul 7, 2016

#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]}

Related questions
See all questions in Polynomial Inequalities
Impact of this question
3088 views around the world
You can reuse this answer
Creative Commons License