It's similar to how you might solve an inequality involving a polynomial. For example,
[tex](x+1)(x-2)>0[/tex]
First consider the case when the left side actually is zero. This happens when [tex]x=-1[/tex] and [tex]x=2[/tex]. Pick numbers that fall to either side of these values and plug them into the inequality.
For instance, take [tex]x=-2[/tex] (to the left of -1), [tex]x=0[/tex] (between -1 and 2), and [tex]x=3[/tex] (to the right of 2). You have
[tex](-2+1)(-2-2)=4>0\implies\text{ all }x<-1\text{ satisfy the inequality}[/tex]
[tex](0+1)(0-2)=-2<0\implies\text{ all }-1<x<2\text{ do not satisfy the inequality}[/tex]
[tex](3+1)(3-2)=4>0\implies\text{ all }x>2\text{ satisfy the inequality}[/tex]
So the polynomial is positive for [tex]x<-1[/tex] and [tex]x>2[/tex].
You can use a similar analysis to determine the sign of the second derivative over certain intervals.
