Question 78550
{{{x^5+24>3x^3+8x^2}}}
{{{x^5-3x^3-8x^2+24>0}}} Get all terms to one side
{{{x^3(x^2-3)-8(x^2-3)>0}}} Factor out the GCF
{{{(x^3-8)(x^2-3)>0}}} Combine like terms
{{{(x^3-8)(x^2-3)=0}}} Now make an equation

Set each factor equal to zero
{{{x^3-8=0}}}
{{{(x-2)(x^2+2x+4)=0}}} use the difference of cubes
Set x-2 equal to zero
{{{x-2=0}}}
{{{x=2}}}
Set {{{x^2+2x+4}}} equal to zero
{{{x^2+2x+4=0}}}
This cannot be factored further


Now set {{{x^2-3}}} equal to zero

{{{x^2-3=0}}}
{{{x^2=3}}}
{{{x=0+-sqrt(3)}}}

So we have

{{{x=sqrt(3)}}} or {{{x=-sqrt(3)}}} or {{{x=2}}}

These are the points where {{{x^5-3x^3-8x^2+24}}} goes through the x-axis. These are the turning points from positive to negative or vice versa. In other words, at these points the graph will change from {{{f(x)>0}}} to {{{f(x)<0}}} or vice versa. So lets graph the equation {{{y=x^5-3x^3-8x^2+24}}}

{{{ graph( 300, 200, -6, 5, -10, 10,x^5-3x^3-8x^2+24) }}}

From the graph we can see that {{{f(x)>0}}} in the interval

[{{{-sqrt(3)}}}, {{{sqrt(3)}}}] U [2,*[Tex \large \infty])