Lesson Explainer: Polynomial Inequalities | Nagwa Lesson Explainer: Polynomial Inequalities | Nagwa

Lesson Explainer: Polynomial Inequalities Mathematics

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 𝑥 that solve (satisfy) one or more inequalities. For instance, 1<𝑥12𝑥72𝑥+22, which is an example of the more general 𝑚<𝑃(𝑥)𝑀, where 𝑃(𝑥) is a polynomial expression and 𝑚 and 𝑀 are numbers. Maybe the best way of seeing what we are being asked is to consider the graph of the polynomial 𝑦=𝑃(𝑥). In the example above, what we get is the curve below.

Note the following:

  1. The curve 𝑦=𝑃(𝑥) here is typical of a cubic polynomial.
  2. The intersection of 𝑦=2 with the curve is the 3 points (𝑏,𝑃(𝑏)),(𝑐,𝑃(𝑐)), and (𝑓,𝑃(𝑓)). Of course, 𝑃(𝑏)=𝑃(𝑐)=𝑃(𝑓)=2. We have labeled the corresponding numbers 𝑏,𝑐, and 𝑓 on the 𝑥-axis.
  3. The points (𝑎,1), (𝑑,1), and (𝑒,1) form the intersection of 𝑦=1 with 𝑦=𝑃(𝑥). Note the numbers 𝑎,𝑑, and 𝑒.
  4. Now if 𝑥]𝑎,𝑏] or 𝑥[𝑐,𝑑[ or 𝑥]𝑒,𝑓], we see that the point (𝑥,𝑃(𝑥)) on the curve lies between lines 𝑦=1 and 𝑦=2, with the line 𝑦=2 included. In other words, 1<𝑃(𝑥)2.

In other words, the polynomial inequality 1<𝑃(𝑥)2 is solved by the union of intervals ]𝑎,𝑏][𝑐,𝑑[]𝑒,𝑓].

To three decimal places, the numbers are actually 𝑎=1.781, 𝑏=1.637, 𝑐=0, 𝑑=0.281, 𝑒=2, and 𝑓=2.137.

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 𝑓(𝑥)0 using the graph below.


From the graph, the solution to the equality 𝑓(𝑥)=0 is the set {4,3}. These are the abscissae of the 𝑥-intercepts.

The inequality 𝑓(𝑥)0 holds when the point (𝑥,𝑓(𝑥)) is on or below the 𝑥-axis. This happens when 𝑥4 or 𝑥3. The solution set is, therefore, ],4][3,[, which is also the complement of the open interval ]4,3[, so it is equivalently ]4,3[.

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

Consider the inequality (𝑥1)(𝑥2)(𝑥3)(𝑥4)>0.

Here, 𝑃(𝑥) is a degree 4 polynomial which is zero at 𝑥=1,2,3, and 4. Any other real number must belong to one of these intervals: 𝐴=],1[,𝐵=]1,2[,𝐶=]2,3[,𝐷=]3,4[,𝐸=]4,[.

It is clear that 𝑃(𝑥) 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 𝑃(𝑥)>0 by selecting any point of the interval, evaluating and checking. A table makes this much easier.


We conclude that 𝑃(𝑥)=(𝑥1)(𝑥2)(𝑥3)(𝑥4)>0 on the union of intervals: 𝐴𝐶𝐸=],1[]2,3[]4,[.

Notice that this was possible because we (i) had a factorization of 𝑃(𝑥) and (ii) we were comparing against 0. Looking for 𝑃(𝑥)>1.3 would require knowing at which points 𝑃(𝑥)=1.3 first. Factoring 𝑃(𝑥) does not help there, although factoring 𝑃(𝑥)1.3 (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 𝑥4, giving your answer using interval notation.


The inequality 𝑥4 has the same solution set as 𝑥40. We factor 𝑥4=(𝑥2)(𝑥+2).

To solve 𝑥40, we need to add the solutions 2 and 2 of 𝑥=4 to the solutions of the strict inequality 𝑥4<0.

For 𝑥4<0, we consider the three intervals 𝐴: <𝑥<2, 𝐵: 2<𝑥<2, and 𝐶: 2<𝑥<.

  • Since 3𝐴 and (3)4=5>0, the interval 𝐴 lies outside the solution set.
  • Since 0𝐵 and (0)4<0, the interval 𝐵 lies inside the solution set.
  • Since 3𝐶 and 34=5>0, the interval 𝐶 lies outside the solution set.

Adding the two places where 𝑥=4 gives us the solution 𝐵{2,2}=]2,2[{2,2}=[2,2] to the inequality 𝑥4.

Example 3: Solving Quadratic Inequalities

Find all solutions to the inequality (𝑥+4)<1369(𝑥+4). Write your answer using intervals.


To rewrite the inequality in the form 𝑃(𝑥)<0, we have the succession of equivalent inequalities: (𝑥+4)<1369(𝑥+4)(𝑥+4)136+9(𝑥+4)<0𝑥+17𝑥84<0(𝑥4)(𝑥+21)<0.

Factoring reveals that we must consider the intervals ],21[, ]21,4[, and ]4,[. 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 𝑥<21, then factor (𝑥+21)<0 while (𝑥4)<0, so the product is positive.
  • If 21<𝑥<4, then (𝑥+21)>0 while (𝑥4)<0, so the product is negative.
  • If 𝑥>4, then (𝑥+21>0) and (𝑥4)>0 and the product is positive.

We conclude that (𝑥4)(𝑥+21)<0 if and only if 𝑥]21,4[.

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

Example 4: Solving Polynomial Inequalities

Given that 𝑓(𝑥)=𝑥3𝑥6𝑥+8 is 10 when 𝑥=3, determine the conditions on 𝑥 so that 𝑓(𝑥)10.


We check that indeed 𝑓(3)=10 and the inequalities become, equivalently, 𝑥3𝑥6𝑥+810𝑥3𝑥6𝑥+8+100𝑥3𝑥6𝑥+180.

The fact that 𝑓(3)=10 means 𝑥3𝑥6𝑥+18=0 when 𝑥=3 and therefore is divisible by (𝑥3). Long division gives us 𝑥3𝑥6𝑥+18=(𝑥3)𝑥6 and we can go further, since 𝑥6=𝑥6𝑥+6, where 62.449.

So for 𝑓(𝑥)>10, we must consider whether 𝑔(𝑥)=𝑥3𝑥6𝑥+18>0 on the intervals ,6,6,6,6,3,]3,[.

This is decided by considering test points: for example, 𝑔(3)=18<0,𝑔(0)=18>0,𝑔(2.5)=0.125<0,𝑔(4)=10>0.

So 𝑓(𝑥)>10 when 𝑥6,6]3,[. Since 𝑓(𝑥)=10 when 𝑥 is one of ±6 or 3, we can write the solution, in terms of conditions on 𝑥, as 6𝑥6𝑥3.or

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 𝐹(𝑥)=𝑥+𝑥5𝑥+3 and that we are told that 𝐹(1)=0. Where is 𝐹(𝑥)>0?

The first step is to factorize 𝐹(𝑥) seeing that (𝑥1) is a factor. We find that 𝐹(𝑥)=(𝑥1)(𝑥+3).

So we consider the intervals ],3[, ]3,1[, and ]1,[. The signs of 𝐹(𝑥) are, respectively, , +, and +. In other words, 𝑥+𝑥5𝑥+3𝑥]3,1[]1,[=]3,[{1}.ifandonlyif

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