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?


1 Answer 1


Sure. Notice how the expression has been factored for you:

$$x^2(x+5)^3(x-3)\ge 0$$

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.

See if you can check the other three regions to finish off the problem. Remember that the solution set must contain the three roots.

  • $\begingroup$ Are there any guidelines for choosing points? You chose -6 which is less than one of the roots, -5. Would you choose something above -5 as well as something below 3 and above 3? $\endgroup$
    – user88528
    Commented Aug 9, 2013 at 0:24
  • 1
    $\begingroup$ @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 $[3,\infty)$. $\endgroup$
    – Jared
    Commented Aug 9, 2013 at 0:27

You must log in to answer this question.