# Solving inequalities at powers greater than $2$

Usually when I see an inequality like $x^2 - 6x - 16 < 0$, I know that the answer is $-2 < x < 8$ because I can picture where the graph would lie below zero. However, for a problem like $x^2(x+5)^3(x-3) \ge 0$, I'm not sure how to set up the inequalities because I can't picture a graph like this as easily. I'm not supposed to use a calculator for this, and I highly doubt my teacher is expecting us to plot points. Is there a trick to figure out inequalities greater than the second power by using the number of exponents given?

$$x^2(x+5)^3(x-3)\ge 0$$
This means that the roots of the polynomial on the left hand side are $-5$, $0$, and $3$, so all of these points are definitely in our solution set. Now, we need to check points from each of the four regions of the real line separated by these roots. Plugging in $-6$ gives us a positive value, so we know that $(-\infty,-5]$ will be a part of the solution set.
• @user2398046: Imagine the real line, with the points $-5$, $0$, and $3$ colored red. Notice how these points split the real line into $4$ regions. To see if a region is in our solution set, we can test any point within that region. I tested the region $(-\infty,5]$ by using the point $-6$. The region $[-5,0]$ can be tested with any number in between $-5$ and $0$, and similarly for the other two regions $[0,3]$ and $[3,\infty)$. – Jared Aug 9 '13 at 0:27