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 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.

Answer

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.

Intervalπ‘₯𝑃(π‘₯)Sign
𝐴024+
𝐡1.5βˆ’0.9375βˆ’
𝐢2.50.5625+
𝐷3.5βˆ’0.9375βˆ’
𝐸524+

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.

Answer

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.

Answer

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.

Answer

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Ÿβˆ’βˆž,βˆ’βˆš6,οŸβˆ’βˆš6,√6,√6,3,]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Ÿβˆ’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

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