In this explainer, we will learn how to solve polynomial inequalities using factoring, end behavior, and sign charts.

In general, we want to find the set of numbers that solve (satisfy) one or more inequalities. For instance, which is an example of the more general where is a polynomial expression and and are numbers. Maybe the best way of seeing what we are being asked is to consider the graph of the polynomial . In the example above, what we get is the curve below.

Note the following:

- The curve here is typical of a cubic polynomial.
- The intersection of with the curve is the 3 points , and . Of course, . We have labeled the corresponding numbers , and on the -axis.
- The points , , and form the intersection of with . Note the numbers , and .
- Now if or or , we see that the point on the curve lies between lines and , with the line included. In other words,

In other words, the polynomial inequality is solved by the union of intervals

To three decimal places, the numbers are actually , , , , , and .

Of course, in general, we will not have access to such a nice graph. In the following examples, we will go through the more practical ways to solve such inequalities.

Our first example does provide a graph.

### Example 1: Solving Quadratic Inequalities Using Graphs

Determine the solution set of the inequality using the graph below.

### Answer

From the graph, the solution to the equality is the set . These are the abscissae of the -intercepts.

The inequality holds when the point is on or below the -axis. This happens when or . The solution set is, therefore, which is also the complement of the open interval , so it is equivalently

One way to solve polynomial inequalities is to factor and then use those factors to decide where the solutions are.

Consider the inequality

Here, is a degree 4 polynomial which is zero at , and 4. Any other real number must belong to one of these intervals:

It is clear that must have the same sign on each of these intervals—if not, it would have to change sign and then have another zero in that interval. This is impossible, since we already listed all the zeros.

So we can tell which interval by selecting any point of the interval, evaluating and checking. A table makes this much easier.

Interval | Sign | ||
---|---|---|---|

0 | 24 | + | |

1.5 | |||

2.5 | 0.5625 | + | |

3.5 | |||

5 | 24 |

We conclude that on the union of intervals:

Notice that this was possible because we (i) had a factorization of and (ii) we were comparing against 0. Looking for would require knowing at which points first. Factoring does not help there, although factoring (if possible) would.

In the absence of a graph, we have the following example.

### Example 2: Solving Quadratic Inequalities

Find all solutions to the inequality , giving your answer using interval notation.

### Answer

The inequality has the same solution set as . We factor

To solve , we need to add the solutions and 2 of to the solutions of the strict inequality .

For , we consider the three intervals : , : , and : .

- Since and , the interval lies outside the solution set.
- Since and , the interval lies inside the solution set.
- Since and , the interval lies outside the solution set.

Adding the two places where gives us the solution to the inequality .

### Example 3: Solving Quadratic Inequalities

Find all solutions to the inequality . Write your answer using intervals.

### Answer

To rewrite the inequality in the form , we have the succession of equivalent inequalities:

Factoring reveals that we must consider the intervals , , and . An alternative to selecting test numbers (which is quicker with quadratic functions like this one) is to think what happens to the factors in each interval:

- If , then factor while , so the product is positive.
- If , then while , so the product is negative.
- If , then and and the product is positive.

We conclude that if and only if .

Here is another example, which should be solved without the aid of a graph.

### Example 4: Solving Polynomial Inequalities

Given that is when , determine the conditions on so that .

### Answer

We check that indeed and the inequalities become, equivalently,

The fact that means when and therefore is divisible by . Long division gives us and we can go further, since where .

So for , we must consider whether on the intervals

This is decided by considering test points: for example,

So when . Since when is one of or 3, we can write the solution, in terms of conditions on , as

You may be tempted to skip the check of each interval with the idea that if the function is negative to one side of a number, it must be positive on the other, and vice versa. This is not always the case!

Consider the function and that we are told that . Where is ?

The first step is to factorize seeing that is a factor. We find that

So we consider the intervals , , and . The signs of are, respectively, , , and . In other words,