for the inequality I meant to write x^3 + 3x < 0
Ah; that makes more sense!
You've got a cubic, which I'm sure you're quite used to graphing. If you think about the graph of y = x3
+ 3x = x(x2
+ 3), you'll recognize that there is one "real" (graphable) solution, namely at x = 0. The quadratic factor is always positive (in fact, always 3 or greater).
So you've got a cubic that crosses the x-axis exactly once. This is a positive cubic (the leading coefficient is an "understood" +1), so the graph comes up from the bottom on the left, crosses the x-axis at x = 0, and then heads up to the right. There may be a "squiggle" in the middle somewhere.
With this zero in mind, what are the two intervals for this inequality? (That is, what are the two parts of the number line where the cubic isn't zero?)
With this graph in mind, on which of those those intervals is the cubic negative? (That is, on which interval is the graph below the x-axis?)
That interval is your solution. Remember to include the point x = 0 in your solution interval, since this is an "or equal to" inequality.