Explainer: Polynomial Inequalities

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 π‘₯ that solve (satisfy) one or more inequalities. For instance, 1<π‘₯βˆ’12π‘₯βˆ’72π‘₯+2≀2, which is an example of the more general π‘š<𝑃(π‘₯)≀𝑀, where 𝑃(π‘₯) is a polynomial expression and π‘š 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(π‘₯). In the example above, what we get is the curve below.

Note the following:

  1. The curve 𝑦=𝑃(π‘₯) 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 π‘₯-axis.
  3. The points (π‘Ž,1), (𝑑,1), and (𝑒,1) form the intersection of 𝑦=1 with 𝑦=𝑃(π‘₯). Note the numbers π‘Ž,𝑑, and 𝑒.
  4. Now if π‘₯∈(π‘Ž,𝑏] or π‘₯∈[𝑐,𝑑) or π‘₯∈(𝑒,𝑓], we see that the point (π‘₯,𝑃(π‘₯)) on the curve lies between lines 𝑦=1 and 𝑦=2, with the line 𝑦=2 included. In other words, 1<𝑃(π‘₯)≀2.

In other words, the polynomial inequality 1<𝑃(π‘₯)≀2 is solved by the union of intervals (π‘Ž,𝑏]βˆͺ[𝑐,𝑑)βˆͺ(𝑒,𝑓].

To three decimal places, the numbers are actually π‘Ž=βˆ’1.781, 𝑏=βˆ’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(π‘₯)≀0 using the graph below.


From the graph, the solution to the equality 𝑓(π‘₯)=0 is the set {βˆ’4,βˆ’3}. These are the abscissae of the π‘₯-intercepts.

The inequality 𝑓(π‘₯)≀0 holds when the point (π‘₯,𝑓(π‘₯)) is on or below the π‘₯-axis. This happens when π‘₯β‰€βˆ’4 or π‘₯β‰₯βˆ’3. The solution set is, therefore, (βˆ’βˆž,βˆ’4]βˆͺ[βˆ’3,∞), which is also the complement of the open interval (βˆ’4,βˆ’3), so it is equivalently β„βˆ’(βˆ’4,βˆ’3).

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

Consider the inequality (π‘₯βˆ’1)(π‘₯βˆ’2)(π‘₯βˆ’3)(π‘₯βˆ’4)>0.

Here, 𝑃(π‘₯) is a degree 4 polynomial which is zero at π‘₯=1,2,3, and 4. Any other real number must belong to one of these intervals: 𝐴=(βˆ’βˆž,1),𝐡=(1,2),𝐢=(2,3),𝐷=(3,4),𝐸=(4,∞).

It is clear that 𝑃(π‘₯) must have the same sign on each of these intervalsβ€”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(π‘₯)>0 by selecting any point of the interval, evaluating and checking. A table makes this much easier.


We conclude that 𝑃(π‘₯)=(π‘₯βˆ’1)(π‘₯βˆ’2)(π‘₯βˆ’3)(π‘₯βˆ’4)>0 on the union of intervals: 𝐴βˆͺ𝐢βˆͺ𝐸=(βˆ’βˆž,1)βˆͺ(2,3)βˆͺ(4,∞).

Notice that this was possible because we (i) had a factorization of 𝑃(π‘₯) and (ii) we were comparing against 0. Looking for 𝑃(π‘₯)>1.3 would require knowing at which points 𝑃(π‘₯)=1.3 first. Factoring 𝑃(π‘₯) does not help there, although factoring 𝑃(π‘₯)βˆ’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 π‘₯≀4, giving your answer using interval notation.


The inequality π‘₯≀4 has the same solution set as π‘₯βˆ’4≀0. We factor π‘₯βˆ’4=(π‘₯βˆ’2)(π‘₯+2).

To solve π‘₯βˆ’4≀0, we need to add the solutions βˆ’2 and 2 of π‘₯=4 to the solutions of the strict inequality π‘₯βˆ’4<0.

For π‘₯βˆ’4<0, we consider the three intervals 𝐴: βˆ’βˆž<π‘₯<βˆ’2, 𝐡: βˆ’2<π‘₯<2, and 𝐢: 2<π‘₯<∞.

  • Since βˆ’3∈𝐴 and (βˆ’3)βˆ’4=5>0, the interval 𝐴 lies outside the solution set.
  • Since 0∈𝐡 and (0)βˆ’4<0, the interval 𝐡 lies inside the solution set.
  • Since 3∈𝐢 and ο€Ή3ο…βˆ’4=5>0, the interval 𝐢 lies outside the solution set.

Adding the two places where π‘₯=4 gives us the solution 𝐡βˆͺ{βˆ’2,2}=(βˆ’2,2)βˆͺ{βˆ’2,2}=[βˆ’2,2] to the inequality π‘₯≀4.

Example 3: Solving Quadratic Inequalities

Find all solutions to the inequality (π‘₯+4)<136βˆ’9(π‘₯+4). Write your answer using intervals.


To rewrite the inequality in the form 𝑃(π‘₯)<0, we have the succession of equivalent inequalities: (π‘₯+4)<136βˆ’9(π‘₯+4)(π‘₯+4)βˆ’136+9(π‘₯+4)<0π‘₯+17π‘₯βˆ’84<0(π‘₯βˆ’4)(π‘₯+21)<0.

Factoring reveals that we must consider the intervals (βˆ’βˆž,21), (βˆ’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 π‘₯<βˆ’21, then factor (π‘₯+21)<0 while (π‘₯βˆ’4)<0, so the product is positive.
  • If βˆ’21<π‘₯<4, then (π‘₯+21)>0 while (π‘₯βˆ’4)<0, so the product is negative.
  • If π‘₯>4, then (π‘₯+21>0) and (π‘₯βˆ’4)>0 and the product is positive.

We conclude that (π‘₯βˆ’4)(π‘₯+21)<0 if and only if π‘₯∈(βˆ’21,4).

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

Example 4: Solving Polynomial Inequalities

Given that 𝑓(π‘₯)=π‘₯βˆ’3π‘₯βˆ’6π‘₯+8 is βˆ’10 when π‘₯=3, determine the conditions on π‘₯ so that 𝑓(π‘₯)β‰₯βˆ’10.


We check that indeed 𝑓(3)=βˆ’10 and the inequalities become, equivalently, π‘₯βˆ’3π‘₯βˆ’6π‘₯+8β‰₯βˆ’10π‘₯βˆ’3π‘₯βˆ’6π‘₯+8+10β‰₯0π‘₯βˆ’3π‘₯βˆ’6π‘₯+18β‰₯0.

The fact that 𝑓(3)=βˆ’10 means π‘₯βˆ’3π‘₯βˆ’6π‘₯+18=0 when π‘₯=3 and therefore is divisible by (π‘₯βˆ’3). Long division gives us π‘₯βˆ’3π‘₯βˆ’6π‘₯+18=(π‘₯βˆ’3)ο€Ήπ‘₯βˆ’6ο…οŠ©οŠ¨οŠ¨ and we can go further, since π‘₯βˆ’6=ο€»π‘₯βˆ’βˆš6π‘₯+√6, where √6β‰ˆ2.449.

So for 𝑓(π‘₯)>βˆ’10, we must consider whether 𝑔(π‘₯)=π‘₯βˆ’3π‘₯βˆ’6π‘₯+18>0 on the intervals ο€»βˆ’βˆž,βˆ’βˆš6,ο€»βˆ’βˆš6,√6,ο€»βˆš6,3,(3,∞).

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

So 𝑓(π‘₯)>βˆ’10 when π‘₯βˆˆο€»βˆ’βˆš6,√6βˆͺ(3,∞). Since 𝑓(π‘₯)=βˆ’10 when π‘₯ is one of ±√6 or 3, we can write the solution, in terms of conditions on π‘₯, as βˆ’βˆš6≀π‘₯β‰€βˆš6π‘₯β‰₯3.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(π‘₯)=π‘₯+π‘₯βˆ’5π‘₯+3 and that we are told that 𝐹(1)=0. Where is 𝐹(π‘₯)>0?

The first step is to factorize 𝐹(π‘₯) seeing that (π‘₯βˆ’1) is a factor. We find that 𝐹(π‘₯)=(π‘₯βˆ’1)(π‘₯+3).

So we consider the intervals (βˆ’βˆž,βˆ’3), (βˆ’3,1), and (1,∞). The signs of 𝐹(π‘₯) are, respectively, βˆ’, +, and +. In other words, π‘₯+π‘₯βˆ’5π‘₯+3π‘₯∈(βˆ’3,1)βˆͺ(1,∞)=(βˆ’3,∞)βˆ’{1}.ifandonlyif

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.