Solving Polynomial Inequalities Lesson

In this lesson, we will learn how to solve a polynomial inequality. We previously learned how to solve a quadratic inequality. The basic idea was to replace the inequality symbol with an equality symbol and then solve the resulting equation. The real number solution(s) to the equation we formed gave us the critical values, which can be used to set up intervals on the number line. We can then test inside of each interval or use a sign chart/table to find our solution. The process for solving a polynomial inequality is exactly the same. When the polynomial has a degree that is greater than 2, the problem becomes more time-consuming. To keep things simple, we will give problems that can be factored easily or presented in factored form. This is typical for this section as it allows us to focus on solving the inequality and not the very tedious process of finding zeros from an unfactored polynomial.

Solving a Polynomial Inequality

$$f(x) < 0$$ Where f(x) is a polynomial function and "<" can be replaced with any inequality symbol.

• Rewrite the inequality so that the right side is 0
  • Write the inequality as:
  • $$f(x) < 0$$
  • "<" can be replaced with the appropriate inequality symbol
• Replace the inequality symbol with an equality symbol and then solve the resulting equation
  • Solve the equation:
  • $$f(x) = 0$$
  • The solutions to the above equation are known as the zeros
  • The real zeros to the above equation correspond to x-intercepts on the graph
  • Non-real complex zeros will not correspond to x-intercepts on the graph
• Use the real zeros to create intervals on the number line
• Use a sign chart/table or test point to find the sign in each interval
• Put the solution together based on the inequality given
  • The solution will contain any region that makes the inequality true
  • Endpoints are included for a non-strict inequality
  • Endpoints are excluded for a strict inequality

The above procedure is made possible by the Intermediate Value Theorem. Recall that the Intermediate Value Theorem tells us if we have a polynomial function f(x), where f(a) and f(b) have opposite signs, then there is at least one value c between a and b where f(c) = 0. We can also say that between two consecutive real zeros, the values of the polynomial function are either all positive (meaning we are above the x-axis) or all negative (meaning we are below the x-axis). For this reason, we can just find our x-intercepts to create intervals on the number line. We can then test inside of each interval to see if the function is positive, meaning it lies above the x-axis, or negative, meaning it lies below the x-axis. Once we have all the information we need, we can then just put the solution together based on the inequality symbol in the problem. Let's look at some examples.
Example #1: Solve each inequality. $$2x^3 - x^2 - 18x + 9 > 0$$ 1) Rewrite the inequality so that the right side is 0.
Here, that is already done for us. $$2x^3 - x^2 - 18x + 9 > 0$$ 2) Replace the inequality symbol with an equality symbol and then solve the resulting equation. $$2x^3 - x^2 - 18x + 9 = 0$$ In this case, we can factor the left side by grouping. $$2x^3 - x^2 - 18x + 9 = 0$$ $$x^2(2x - 1) - 9(2x - 1) = 0$$ $$(2x - 1)(x^2 - 9) = 0$$ $$(2x - 1)(x + 3)(x - 3) = 0$$ We can solve this equation using the zero-product property, however, the easier way is to write each factor in the form of (x - k). $$2\left(x - \frac{1}{2}\right)(x - (-3))(x - 3) = 0$$ This tells us we will have zeros of 1/2, -3, and 3. So the x-intercepts will be: $$(-3, 0), \left(\frac{1}{2}, 0\right), (3, 0)$$ To make the problem simpler, we can divide both sides of the inequality by 2. $$\left(x - \frac{1}{2}\right)(x - (-3))(x - 3) > 0$$ 3) Use the real zeros to set up intervals on the number line. We will have four intervals. $$(-∞, -3), \left(-3, \frac{1}{2}\right), \left(\frac{1}{2}, 3\right), (3, ∞)$$ 4) Let's make a sign table, you can also use a test point in each interval. 5) To write our solution, we find that according to our sign table above, the polynomial on the left side of our inequality will be positive for x-values between -3 and 1/2, and then again for x-values that are greater than 3. Since we have a strict inequality, we can report our answer as: $$-3 < x < \frac{1}{2}$$ $$\text{or}$$ $$x > 3$$ Interval Notation: $$\left(-3, \frac{1}{2}\right) ∪ (3, ∞)$$ Graphing the Solution on the Number Line: It can be useful to use a graph to verify your solution. This is where graphing software would really come in handy as creating a graph is very time-consuming.
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As we can see from the graph above, for x-values between -3 and 1/2, the y-values are positive (since we are above the x-axis). Additionally, this also occurs for x-values greater than 3. This confirms our given solution.
Example #2: Solve each inequality. $$x^4 - 6x^3 + 7x^2 + 6x - 8 ≤ 0$$ 1) Rewrite the inequality so that the right side is 0.
Here, that is already done for us. $$x^4 - 6x^3 + 7x^2 + 6x - 8 ≤ 0$$ 2) Replace the inequality symbol with an equality symbol and then solve the resulting equation. $$x^4 - 6x^3 + 7x^2 + 6x - 8 = 0$$ In this case, we can factor by creating a group that is quadratic in form and then another group with what is left over. $$(x^4 + 7x^2 - 8) + (-6x^3 + 6x) = 0$$ $$(x^2 - 1)(x^2 + 8)- 6x(x^2 - 1) = 0$$ $$(x^2 - 1)(x^2 - 6x + 8) = 0$$ $$(x + 1)(x - 1)(x - 4)(x - 2) = 0$$ We can solve this equation using the zero-product property, however, the easier way is to write each factor in the form of (x - k). $$(x - (-1))(x - 1)(x - 4)(x - 2) = 0$$ This tells us we will have zeros of -1, 1, 2, and 4. So the x-intercepts will be: $$(-1, 0), (1, 0), (2, 0), (4, 0)$$ 3) Use the real zeros to set up intervals on the number line. We will have five intervals. $$(-∞, -1), (-1, 1), (1, 2), (2, 4), (4, ∞)$$ 4) Let's make a sign table, you can also use a test point in each interval. 5) To write our solution, we find that according to our sign table above, the polynomial on the left side of our inequality will be negative for x-values between -1 and 1, and then again for x-values that are between 2 and 4. Since we have a non-strict inequality, we can report our answer as: $$-1 ≤ x ≤ 1$$ $$\text{or}$$ $$2 ≤ x ≤ 4$$ Interval Notation: $$[-1, 1] ∪ [2, 4]$$ Graphing the Solution on the Number Line: Again, we can confirm our answer using graphing software.
Desmos Link for More Detail
As we can see from the graph above, for x-values between -1 and 1, the y-values are negative (since we are below the x-axis). Additionally, this also occurs for x-values between 2 and 4. The x-values of -1, 1, 2, and 4 are included since we have a non-strict inequality. This confirms our given solution.

Solving Polynomial Inequalities with Non-Real Complex Zeros

In some cases, we will have non-real complex zeros. These will not show up as x-intercepts on the graph. Let's look at an example.
Example #3: Solve each inequality. $$(x - 2)(2x + 7)(x^2 + x + 1) ≤ 0$$ 1) Rewrite the inequality so that the right side is 0.
Here, that is already done for us. $$(x - 2)(2x + 7)(x^2 + x + 1) ≤ 0$$ 2) Replace the inequality symbol with an equality symbol and then solve the resulting equation. $$(x - 2)(2x + 7)(x^2 + x + 1) = 0$$ As we have done on previous problems, we can rewrite the left side to find our zeros. $$2(x - 2)\left(x - \left(-\frac{7}{2}\right)\right)(x^2 + x + 1) = 0$$ At this point, we can identify that 2 and -7/2 are zeros. Notice that x² + x + 1 is an irreducible quadratic since it can't be factored using real numbers. Let's set this factor equal to zero and solve using the quadratic formula. $$x^2 + x + 1 = 0$$ $$a = 1, b = 1, c = 1$$ $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)}$$ $$x = \frac{-1 \pm i \sqrt{3}}{2}$$ So these are non-real complex zeros. They will not show up as x-intercepts and we can ignore these zeros when setting up intervals on the number line. $$2(x - 2)\left(x - \left(-\frac{7}{2}\right)\right)(x^2 + x + 1) ≤ 0$$ Let's divide each side by 2 to simplify the problem. $$(x - 2)\left(x - \left(-\frac{7}{2}\right)\right)(x^2 + x + 1) ≤ 0$$ The real zeros are 2 and -7/2. So the x-intercepts will be:$$\left(-\frac{7}{2}, 0\right), (2, 0)$$ 3) Use the real zeros to set up intervals on the number line. We will have three intervals. $$\left(-∞, -\frac{7}{2}\right), \left(-\frac{7}{2}, 2\right), (2, ∞)$$ 4) Let's make a sign table, you can also use a test point in each interval. Let's stop for a moment and answer a question you may have. $$x^2 + x + 1 > 0$$ The answer for the above is all real numbers. How do we know this? We have non-real complex roots for the left side of the inequality. This means the graph will never touch the x-axis. So the function values are either always positive or always negative for any real number. When this occurs, you can pick any real number you want and just plug in. Let's try an x-value of 0. $$f(0) = 0^2 + 0 + 1 = 1$$ This means that x² + x + 1 is always positive. It will never be zero or negative. We can easily confirm this by looking at graph.
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Notice from the graph above that our function values are never zero or negative. We are always above the x-axis where function values are positive.
5) To write our solution, we find that according to our sign table above, the polynomial on the left side of our inequality will be negative for x-values between -7/2 and 2. Since we have a non-strict inequality, we can report our answer as: $$-\frac{7}{2} ≤ x ≤ 2$$ Interval Notation: $$\left[-\frac{7}{2}, 2\right]$$ Graphing the Solution on the Number Line: Again, we can confirm our answer using graphing software.
Desmos Link for More Detail
As we can see from the graph above, for x-values between -7/2 and 2, the y-values are negative (since we are below the x-axis). The x-values of -7/2 and 2 are included since we have a non-strict inequality. This confirms our given solution.

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