It's similar to how you might solve an inequality involving a polynomial. For example,
\((x+1)(x-2)>0\)
First consider the case when the left side actually is zero. This happens when \(x=-1\) and \(x=2\). Pick numbers that fall to either side of these values and plug them into the inequality.
For instance, take \(x=-2\) (to the left of -1), \(x=0\) (between -1 and 2), and \(x=3\) (to the right of 2). You have
\((-2+1)(-2-2)=4>0\implies\text{ all }x<-1\text{ satisfy the inequality}\)
\((0+1)(0-2)=-2<0\implies\text{ all }-1
\((3+1)(3-2)=4>0\implies\text{ all }x>2\text{ satisfy the inequality}\)
So the polynomial is positive for \(x<-1\) and \(x>2\).
You can use a similar analysis to determine the sign of the second derivative over certain intervals.
