Section 2.7: Polynomial and Rational Inequalities

Learning Outcomes

Solve polynomial inequalities algebraically.
Solve rational inequalities algebraically.

Solving Polynomial Inequalities

One application of our ability to find intercepts and sketch a graph of polynomials is the ability to solve polynomial inequalities. It is a very common question to ask when a function will be positive and negative.

When we solve an equation and the result is $x=3$ , we know there is one solution, which is 3.

When we solve an inequality and the result is $x<3$ , we know there are many solutions. We graph the result to better help show all the solutions, and we start with 3. Three becomes a critical point and then we decide whether to shade to the left or right of it. The numbers to the right of 3 are larger than 3, so we shade to the right.
Number line of x greater than 3
To solve a polynomial inequality, we must write the inequality with the polynomial on the left and 0 on the right.

Next, we determine the critical points to use to divide the number line into intervals. A critical point is a number which makes the polynomial zero.

We will evaluate the factors of the polynomial. This will identify the interval, or intervals, that contains all the solutions of the polynomial inequality.

We write the solution in interval notation being careful to determine whether the endpoints are included.

How To: Solve a polynomial inequality

1. Write the inequality with the polynomial on the left and zero on the right
2. Determine the critical points-the points where the polynomial will be zero.
3. Use the critical points to divide the number line into intervals.
4. Test a value in each interval. Indicate which regions are positive and negative.
5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

Example 1: Solving A Polynomial InequalitY in Factored FORM

Solve the inequality $\left(x+3\right){\left(x+1\right)}^{2}\left(x-4\right)> 0$

Show Solution

As with all inequalities, we start by solving the equality $\left(x+3\right){\left(x+1\right)}^{2}\left(x-4\right)= 0$ , which has solutions at x = -3, -1, and 4. We know the function can only change from positive to negative at these values, so these divide the inputs into 4 intervals.
We could choose a test value in each interval and evaluate the function $f\left(x\right) = \left(x+3\right){\left(x+1\right)}^{2}\left(x-4\right)$ at each test value to determine if the function is positive or negative in that interval

Interval	Test x in interval	f(test value)	> 0 or < 0
x < -3	-4	72	> 0
-3 < x < -1	-2	-6	< 0
-1 < x < 4	0	-12	< 0
x > 4	5	288	> 0

On a number line this would look like:

Number line with values from -6 to 6 double headed arrows from -6 to -3 read positive, from -3 to -1 read negative, from -1 to positive 4 read negative and from 4 to 6 read positive.
From our test values, we can determine this function is positive when x < -3 or x > 4, or in interval notation, $\left(-\infty, -3\right)\cup\left(4,\infty\right)$ . We could have also determined on which intervals the function was positive by sketching a graph of the function. We illustrate that technique in the next example.

Example 2: Solving A Quadratic InequalitY NOT in Factored FORM

Solve the inequality $6-5t-{t}^{2}\ge 0$

Show Solution

As with all inequalities, we start by solving the equality $6 - 5t - {t}^{2}= 0$ . The first step is to write this in factored form. While we could use the quadratic formula, this equation factors nicely to $\left(6 + t\right)\left(1-t\right)=0$ , giving horizontal intercepts t = 1 and t = -6. We know the function can only change from positive to negative at these values, so these divide the inputs into 3 intervals.

We could choose a test value in each interval and evaluate the function $f\left(t\right) = 6-5t-{t}^{2} = 0$ at each test value to determine if the function is positive or negative in that interval

Interval	Test t in interval	f(test value)	> 0 or < 0
t < -6	-7	-8	< 0
-6 < t < 1	0	6	> 0
t > 1	2	-8	< 0

From our test values, we can determine this function is positive when -6 < t < 1, or in interval notation, $\left[-6,1\right]$ .

Example 3: Solving a Polynomial Inequality Not in Factored Form

Solve the inequality ${x}^{4} - 2{x}^{3} - 3{x}^{2} \gt 0$

Show Solution

In our other examples, we were given polynomials that were already in factored form, here we have an additional step to finding the intervals on which solutions to the given inequality lie. Again, we will start by solving the equality ${x}^{4} - 2{x}^{3} - 3{x}^{2} = 0$

Notice that there is a common factor of ${x}^{2}$ in each term of this polynomial. We can use factoring to simplify in the following way:

$\begin{align}{x}^{4} - 2{x}^{3} - 3{x}^{2} &= 0&\\{x}^{2}\left({x}^{2} - 2{x} - 3\right) &= 0\\ {x}^{2}\left(x - 3\right)\left(x + 1 \right)&= 0\end{align}$

Now we can set each factor equal to zero to find the solution to the equality.

$\begin{array}{ccc} {x}^{2} = 0 & \left(x - 3\right) = 0 &\left(x+1\right) = 0\\ {x} = 0 & x = 3 & x = -1\\ \end{array}$ .

Note that x = 0 has multiplicity of two, but since our inequality is strictly greater than, we don’t need to include it in our solutions.
We can choose a test value in each interval and evaluate the function

${x}^{4} - 2{x}^{3} - 3{x}^{2} = 0$

at each test value to determine if the function is positive or negative in that interval

Interval	Test x in interval	> 0, < 0
x < -1	-2	x > 0
-1 < x < 0	-1/2	x < 0
0 < x < 3	1	x < 0
x > 3	5	x > 0

We want to have the set of x values that will give us the intervals where the polynomial is greater than zero. Our answer will be $\left(-\infty, -1\right]\cup\left[3,\infty\right)$ .

The graph of the function gives us additional confirmation of our solution.

Line dips down, dips slightly up, dips very far down, then sharply goes up

Try it

Solving Rational Inequalities

In addition to finding when a polynomial function will be positive and negative, we can also find where rational functions are positive and negative.

A rational function is a function that can be written as the quotient of two polynomial functions. Rational functions will be studied in more detail in the next section.

A General Note: Rational Function

A rational function is a function that can be written as the quotient of two polynomial functions $P\left(x\right) \text{and} Q\left(x\right)$ .

$f\left(x\right)=\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{{a}_{p}{x}^{p}+{a}_{p - 1}{x}^{p - 1}+...+{a}_{1}x+{a}_{0}}{{b}_{q}{x}^{q}+{b}_{q - 1}{x}^{q - 1}+...+{b}_{1}x+{b}_{0}},Q\left(x\right)\ne 0$

Rational Inequality

A rational inequality is an inequality that contains a rational expression.

Inequalities such as $\frac{3}{2x}>1\text{ , }\frac{2x}{x-3}<4\text{ , }\frac{(2x-3)(x-5)}{x+1)^{2}}<0\text{ , and }\frac{1}{4}-\frac{2}{x^{2}}\leq\frac{3}{x}$ are rational inequalities as they each contain a rational expression. When we solve a rational inequality, we will use the same techniques we used solving polynomial inequalities. However, we must carefully consider what value might make the rational expression undefined and so must be excluded. Next, we determine the critical points to use to divide the number line into intervals. A critical point is a number which makes the rational expression zero of undefined.

We will evaluate the factors of the numerator and denominator, and find the quotient in each interval. This will identify the interval, or intervals, that contains all the solutions of the rational inequality.

We write the solution in interval notation being careful to determine whether the endpoints are included.

How To: Solve a rational inequality

1. Write the inequality as one quotient on the left and zero on the right
2. Determine the critical points-the points where the rational expression will be zero or undefined
3. Use the critical points to divide the number line into intervals.
4. Test a value in each interval. Indicate which regions are positive and negative.
5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

Example 4: Solving RATIONAL Inequalities in Factored Form

Solve $\dfrac{(x+3)(x-4)}{(x+1)^{2}}\leq 0$

Show Solution

As with all inequalities, we start by solving the equality $\dfrac{(x+3)(x-4)}{(x+1)^{2}}= 0$ , which has solutions at x = -3, -1, and 4. We know the function can only change from positive to negative at these values, so these divide the inputs into 4 intervals.
We could choose a test value in each interval and evaluate the function $f\left(x\right) = \dfrac{(x+3)(x-4)}{(x+1)^{2}}= 0$ at each test value to determine if the function is positive or negative in that interval

Interval	Test x in interval	f(test value)	> 0 or < 0
x < -3	-4	0.888	> 0
-3 < x < -1	-2	-6	< 0
-1 < x < 4	0	-12	< 0
x > 4	5	0.222	> 0

On a number line this would look like:

Example 5: Solving A rational inequalitY not in factored form

Solve $\dfrac{2x^{2}+6x+9}{x+3}>5$

Show Solution

This time, the rational inequality is not in factored form, and there is not a zero on the right side. We need to subtract the 5 to the other side and get common denominators: $\dfrac{2x^{2}+6x+9}{x+3}-5>0 \\ \dfrac{2x^{2}+6x+9}{x+3}-\dfrac{5(x+3)}{x+3}>0 \\ \dfrac{2x^{2}+6x+9-5(x+3)}{x+3}>0 \\ \dfrac{2x^{2}+x-6}{x+3}>0 \\ \dfrac{(x+2)(2x-3)}{x+3}>0$

We will now solve $\dfrac{(x+2)(2x-3)}{x+3}=0$ which has solutions at $x=-2,\dfrac{3}{2},-3$ . We know the function can only change from positive to negative at these values, so these divide the inputs into 4 intervals.
We could choose a test value in each interval and evaluate the function $f\left(x\right) = \dfrac{(x+2)(2x-3)}{x+3}$ at each test value to determine if the function is positive or negative in that interval

Interval	Test x in interval	f(test value)	> 0 or < 0
x < -3	-4	-22	< 0
-3 < x < -2	-2.5	8	> 0
-2 < x < 3/2	0	-2	< 0
x > 3/2	2	0.8	> 0

From our test values, we can determine this function is positive when [latex]-3\dfrac{3}{2}[/latex], or in interval notation: $(-3,-2) \cup \left(\dfrac{3}{2},\infty \right)$ .

Try it

Solve $\dfrac{(x-6)(x+1)}{3x^{2}}<0$ and write your answer in interval notation.

Show Solution

$(-1,0)\cup(0,6)$

Section 2.7 Homework Exercises

1. Explain critical points and how they are used to solve polynomial and rational inequalities algebraically.

2. Describe the steps needed to solve a rational inequality algebraically.

For each of the following polynomial inequalities, solve and write your answer in interval notation

3. $\text{ }(x-4)(x+3)\lt 0$

4. $\text{ }(x-4)(x+1)\lt 0$

5. $\text{ }(x-1)(3x-4)\geq 0$

6. $\text{ }(x+7)(2x-5)\geq 0$

7. $\text{ }x^{2}+5x+4\lt 0$

8. $\text{ }x^{2}-14x+49\lt 0$

9. $\text{ }(x+8)(x+2)(x-3)\geq 0$

10. $\text{ }(x+5)(x+1)(x-4)\geq 0$

11. $\text{ }(x+8)^{2}(x+5)(x+7)^{3}\gt 0$

12. $\text{ }(x-4)(x+1)^{3}(x-2)^{2}\gt 0$

13. $\text{ }\left(2x^{2}+5x-3\right)\left(x+4\right)\leq 0$

14. $\text{ }\left(3x^{2}-5x-2\right)\left(x+3\right)\leq 0$

15. $\text{ }\left(x^{2}+3x-10\right)\left(x^{2}-1\right)\lt 0$

16. $\text{ }\left(x^{2}+2x-24\right)\left(x^{2}-4\right)\lt 0$

For each of the following rational inequalities, solve and write your answer in interval notation.

17. $\text{ }\dfrac{x-7}{x-1}\lt 0$

18. $\text{ }\dfrac{x+5}{x-4}\lt 0$

19. $\text{ }\dfrac{x+4}{2x-3}\geq 0$

20. $\text{ }\dfrac{x-5}{3x+4}\geq 0$

21. $\text{ }\dfrac{x+32}{x+6}\geq 3$

22. $\text{ }\dfrac{x+68}{x+8}\geq 5$

23. $\text{ }\dfrac{(x+3)(x+5)}{x+2}\geq 0$

24. $\text{ }\dfrac{(x+4)(x-3)}{x+1}\geq 0$

25. $\text{ }\dfrac{(x-2)(x-7)}{3x+2}\leq 0$

26. $\text{ }\dfrac{(x+1)(x+6)}{2x-1}\leq 0$

27. $\text{ }\dfrac{x+6}{x^{2}-5x-24}\leq 0$

28. $\text{ }\dfrac{x+5}{x^{2}-9x+18}\leq 0$

29. $\text{ }\dfrac{(x+7)(x-3)}{(x-5)^{2}}\lt 0$

30. $\text{ }\dfrac{(x+8)(x-1)}{(x+3)^{2}}\lt 0$

31. $\text{ }\dfrac{2x^{2}+6x+7}{x+3}\gt 3$

32. $\text{ }\dfrac{4x^{2}+8x+9}{x+2}\gt 5$

Chapter 2: Polynomial and Rational Functions