Examples of Solving Rational Inequalities

This is already factored for me, so I don't have to bother with that. I'll go straight to finding the zeroes (from the numerator) and the undefined points (from the denominator):

So the number line is split into the intervals (−∞, −5), (−5, 2), (2, 3), and (3, +∞). Now I'll find where each factor is positive:

The negative factor, −2x − 10, and the backwards factor, 3 − x, have given me backwards inequalities, so the factor table looks like this:

(The x − 2 factor is listed twice, because that factor occurs twice. If that factor had been cubed, it would have been listed in the table three times. The x² + 5 factor has all "plus" signs in its row, because this factor is never zero or negative.)

Looking at the signs in the bottom row, I see that the rational expression is negative on the intervals (−5, 2) and (2, 3). Since this problem is not an "or equal to" inequality, I don't need to consider the endpoints; I know they don't belong in the solution. But I do need to remember that x = 2 is not part of the solution, and I must resist the impulse to join these two intervals together (by throwing in x = 2) to get an incorrect solution interval of (−5, 3).

Since I can't include x = 2, the solution is two separate intervals:

Before I can start working with this inequality, I have to gather the two rational expressions on one side of the inequality symbol, and combine them:

I tried to factor the quadratic in the denominator, but it's unfactorable (that is, it's prime). So I ran it through the Quadratic Formula. The zeroes turned out to be complex numbers, so it has no real-number zeroes; in particular, the quadratic's parabola never crosses the x-axis. This is a negative quadratic, which means that it opens downward; with no real zeroes, the graph is always going to be below the x-axis, so this factor is always negative.

This leaves me with the two linear factors. The factor in the numerator *can* include the endpoint where it changes signs:

For the denominator's linear factor, I cannot include the endpoint, because this would cause division by zero:

Now I can do my factor table:

I recall that I cannot include 1 within the solution, because this would cause division by zero. But I am allowed to include the other endpoint. So my answer is: