Question 258043
{{{3x^3+4x^2-15x<0}}}
At the start we solve inequalities like this the same way you solve equations: Get one side equal to zero and factor. The right side is already zero so we will start by factoring. When factoring, always start with the Greatest Common Factor (GCF). The GCF here is x. Facroting out x we get:
{{{x(3x^2+4x-15)<0}}}
Next we factor the trinomial:
{{{x(3x-5)(x+3)<0}}}<br>
Next let's find the values for x that make the factors zero. Setting each factor to zero and solving we get:
x = 0 or x = 5/3 or x = -3, respectively
Since the equality says "less than zero" and not "less than or equal to zero", these x values are not part of our solution. But they will help us figure out the  answer. Since these x values are not part of the solution, plot these numbers as open circles on a number line. (Unfortunately Algebra.com's graphing soltware does not graph inequalities so I will be forced to describe the graph.)<br>
These three open circles divide the number line into 4 sections.
Let's analyze what happens to our three factors for the x values in each section:<ul><li>To the left of -3. In this section all three factors are negative. And the product of 3 negative numbers is negative. So this section is part of our solution. Graph a shaded line starting from the open circle at -3 and going left. Put an arrowhead on the end of the shaded line to indicate that it continues on to the left forever.</li><li>Between -3 and 0. The x+3 factor is positive but the other two are still negative. The product of one positive and two negative numbers is positive. This section is not part of our solution.</li><li>Between 0 and 5/3. The x+3 and x factors are positive and the 3x-5 factor is still negative. The product of two positive and one negative number is negative. This section is part of our solution. Shade the number line between the open circles at 0 and 5/3. (No arrowheads on this shaded area.)</li><li>To the right of 5/3. Here all three factors are positive. The product of three positive numbers is positive. This is not part of the solution to our inequality.</li></ul>
In summary our graph has
1) A shaded arrow starting from an open circle at -3 and going left.
2) A shaded section of the number line between open circles at 0 and 5/3.