Solving rational inequalities is very similar to solving polynomial inequalities.

But because rational expressions have denominators (and therefore may have places where they're not defined), you have to be a little more careful in finding your solutions.

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To solve a rational inequality, use these steps:

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- If needed, move all the terms to one side of the inequality symbol, with zero on the other side.
- If needed, combine all rational expressions into one polynomial fraction.
- Factor the numerator and denominator completely.
- Solve the factors for their zeroes; keep in mind that the denominator's zeroes would cause division by zero, so they cannot be included in your solution.
- Use the zeroes to divide the number line into intervals.
- Make a table of factors, showing where each factor is less than and greater than zero.
- Multiply the factors' signs on each interval (that is, down the table's columns) to find the sign of the rational expression on that interval.
- Select the intervals that match the inequality they gave you; remember to discard any endpoints that would cause division by zero.

Let's see how these instructions work in practice:

- Solve the following:

They've already put this inequality into (one rational expression) with (zero) on the other side. So I can start with factoring everything:

This polynomial fraction will be zero wherever the numerator is zero, so I'll set the numerator equal to zero and solve:

(*x* + 2)(*x* + 1) = 0

*x* + 2 = 0 or *x* + 1 = 0

*x* = −2 or *x* = −1

The fraction will be undefined wherever the denominator is zero, so I'll set the denominator equal to zero and solve:

(*x* + 4)(*x* − 4) = 0

*x* + 4 = 0 or *x* − 4 = 0

*x* = −4 or *x* = 4

These four values, −4, −2, −1, and +4, divide the number line into five intervals, namely:

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(−∞, −4)

(−4, −2)

(−2, −1)

(−1, 4)

(4, +∞)

I could use "test points" to find the solution to the inequality, by picking an *x*-value in each interval, plugging it into the original rational expression, simplifying to get a numerical answer, and then checking the sign, but that process gets long and annoying (and is prone to errors), so I'll use the easier and faster factor-table method instead.

My factor table looks like this:

My table has one row for each factor, a row for the number line, and a row for the rational expression. Each row is split into columns, with each column corresponding to one of the intervals on the number line.

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The sign of the overall rational expression is a result of the signs of its various factors, so I need to find where each factor is positive:

*x* + 4 > 0 for *x* > −4
*x* + 2 > 0 for *x* > −2
*x* + 1 > 0 for *x* > −1
*x* − 4 > 0 for *x* > 4

Now I can put "plus" signs on the intervals in each row where that row's factor is positive:

Wherever a factor isn't positive, it's negative, so I'll put "minus" signs in the remaining entries of each row:

I know that the product of an even number of "minus" signs is a plus; the product of an odd number of "minus" signs is a minus. So, by multiplying the signs down the columns (or just counting up the minusses), I get the overall sign of the rational expression on each interval:

Then the rational is positive on the intervals (−∞, −4), (−2, −1), and (4, +∞).

Looking back at the original exercise, this is an "or equal to" inequality, so I need to consider the interval endpoints, too.

If this were a polynomial inequality, I could just throw all the interval endpoints into the solution, and I'd be done. For rational expressions, though, I have to be careful not to include any *x*-values that would cause division by zero.

The intervals' endpoints are −4, −2, −1, and 4. I can include −2 and −1 in the solution, because they just make the expression equal to zero by making the numerator zero. But plugging −4 or 4 into the rational expression would cause division by zero, making the rational expression undefined, so I can't include these values in the solution.

Then my full solution is:

I wrote my solution above in "interval" notation. If you have to write your solution in "inequality" notation, it would look like this:

*x* < −4, −2 ≤ *x* ≤ −1, and *x* > 4

Don't forget: "Infinity" is not a "number" in the way that, say, "2" is. "Infinity" cannot be "included" in your solution, so never draw a square bracket next to an "infinity" "endpoint".

URL: https://www.purplemath.com/modules/ineqrtnl.htm

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