4.4: Polynomial Inequalities
- Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Solving Polynomial Inequalities
How can we solve inequalities involving polynomials such as \(x^3-4x^2+3x > 0 \)? Determining where \(x^3-4x^2+3x > 0 \) is the same as determining where the function \(y=f(x) = x^3-4x^2+3x \) is positive. Similarly, determining where \(x^3-4x^2+3x < 0 \) is the same as determining where the function \(y=f(x) = x^3-4x^2+3x \) is negative. One approach that we can utilize is to use the graph of the polynomial to determine the sign of the polynomial using the zeros and the end behavior. When the graph is above the x-axis, the function is positive. When the graph is below the x-axis, the function is negative. When the graph is on the x-axis, the function value is zero.
Solve \(x^3-4x^2+3x > 0 \)
Solution
We can solve this inequality by determining where \(y=f(x) = x^3-4x^2+3x \) is positive. Factoring, we have \[f(x) = x^3-4x^2+3x = x(x^2-4x+3) = x(x-3)(x-1) \nonumber \] So the zeros are \[x=0, x=3, \text{ and } x=1 \nonumber \] each with multiplicity one. Therefore, the graph will cross the x-axis at each zero. The end behavior of \(f\) is the same as \(y=x^3\) which opens up on the right (as \(x \to \infty\), \(y \to \infty\)) and down on the left (as \(x \to -\infty\), \(y \to -\infty\)). Using the zeros and multiplicity, we can graph \(y=f(x)\) as shown below.
From the graph, we can see that \(f\) is positive on \( (0,1) \) and \( (3 , \infty) \) when the graph is above the x-axis. These intervals form the solution to our inequality \( (0,1) \cup (3, \infty) \)
Solve \(x^4-x^3-6x^2 < 0 \)
Solution
We can solve this inequality by determining where \(y=f(x) = x^4-x^3-6x^2 < 0 \) is negative or zero. Factoring, we have \[f(x) = x^4-x^3-6x^2 = x^2(x^2-x-6) = x^2(x-3)(x+2) \nonumber \] So the zeros are \(x=3\) with multiplicity one, \(x=0\) with multiplicity two, and \(x=-2\) with multiplicity one. Therefore, the graph will cross the x-axis at \(x=3\) and \(x=-2\) but will not cross the x-axis at \(x=0\) . The end behavior of \(f\) is the same as \(y=x^4\), an even power, which opens up on both ends (as \(x \to \pm \infty\), \(y \to \infty\)). Using the zeros and multiplicity, we can graph \(y=f(x)\) as shown below.
From the graph, we can see that \(f\) is negative on \( (-2,0) \cup (0,3) \) when the graph is below the x-axis. This interval form the solution to our inequality \( (-2,0) \cup (0,3) \).
What is different about the inequality below? \[x^3+2x^2 < -5x \nonumber \] The expression \(-5x \) could be positive or negative depending on the value of \(x\), so we can't make this a question of where a function is positive or negative. However, if we get zero on one side of the inequality by adding \(5x\) to both sides, we have \[x^3+2x^2 +5x < 0 \nonumber \] This inequality can be solved by determining where \(y=f(x)=x^3+2x^2 +5x \) is negative as shown in example 4.4.3 below.
A polynomial inequality of the form \(f(x) > 0\) or \(f(x) < 0\) can be solved by determining the sign of \(f(x)\) using the graph.
Solve \(x^3+2x^2 < -5x\)
Solution
First, getting zero on once side of the inequality \[x^3+2x^2 +5x < 0 \nonumber \] We can solve this inequality by determining where \(y=f(x) = x^3+2x^2 +5x \) is negative. Factoring, we have \[f(x) = x^3+2x^2 +5x = x(x^2+2x+5) \nonumber \] Setting each factor to zero, we have \[ x=0 \text{ and } x^2+2x+5=0 \nonumber \] The first equation gives the zero \(x=0\) with multiplicity one. Using the quadratic formula to solve the second equation with \(a=1\), \(b=2\), and \(c=5\), we have \[x= \dfrac{-2 \pm \sqrt{2^2-4(1)(5)}}{2(1)}\nonumber \]\[x= \dfrac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i \nonumber \] Since these are a complex zeros, the graph will only cross the x-axis at the real zero \(x=0\). The end behavior of \(f\) is the same as \(y=x^3\) which opens up on the right (as \(x \to \infty\), \(y \to \infty\)) and down on the left (as \(x \to -\infty\), \(y \to -\infty\)). Using the zeros and multiplicity, we can graph \(y=f(x)\) as shown below.
From the graph, we can see that \(f\) is negative on \( (-\infty,0) \) when the graph is below the x-axis. This interval form the solution to our inequality \( (-\infty,0) \).
Note that if the inequality in Example 4.4.3 was altered to \(x^3+2x^2 \le -5x\), this would change our solution slightly. The inequality is equivalent to \(x^3+2x^2 +5x \le 0\). The solutions are values of \(x\) where \(f(x)=x^3+2x^2 +5x\) is negative or zero. We would have to include the real zero \(x=0\) in our solution in this case. Our solution is \( (-\infty,0] \).
Solve \(x^3-7x^2 \ge -12x\)
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The inequality is equivalent to \[x^3 -7x^2 +12x \ge 0 \nonumber \] To solve this inequality determine where \(y=f(x) = x^3 -7x^2 +12x \) is positive, using the zeros \(x=0\), \(x=3\), and \(x=4\) (each with multiplicity one) and the end behavior (resembling that of \(y=x^3\)). The graph below shows that \(f\) is positive or zero on \( [3,0] \cup [4, \infty) \). These intervals are the solution to the inequality.
Explain what is wrong with the following solution to the inequality \[x^2-2x > 0 \nonumber \]\[x(x-2) > 0 \nonumber \] \[x > 0 \text{ or } (x-2) > 0 \nonumber \] \[x > 0 \text{ and } x > 2 \nonumber \] The solutions to the inequality is \( (0, \infty) \cup (2, \infty) \).
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There are two problems with the solution. An error occurs in the third step. Just because two factors multiply to zero in step two \( x(x-2) \ge 0 \) does not necessarily mean that each factor must be greater that zero or positive as shown in step three where \(x \ge 0 \text{ and } (x-2) \ge 0 \). The two factors on the left side of the inequality could either be both positive or both negative for the product to be positive. This solution incorrectly reflects how we should solve the equation \(x^2-2x = 0 \) as opposed to an inequality. Factoring \[x(x-2) = 0 \nonumber \]Setting each factor to zero \[x = 0 \text{ or } x-2=0 \nonumber \]Then solving for x, we have \[x = 0 \text{ or } x = 2 \nonumber \]This is the correct way to solve an equality not an inequality.
A second issue occurs in the last step. The values of x where \(x > 0 \text{ or } x > 2 \) are merely the interval the values of x where \(x > 0 \). If \(x > 2 \), then \(x > 0 \) also. The second inequality is redundant.
Focus: Another Approach to Solving Polynomial Inequalities
Notice on the graph in Example 4.3.2, the function \(f(x)=x^4-x^3-6x^2\) only changes sign from positive to negative (or vice versa) at the zeros \(x=3\) and \(x=2\). Although \(x=0\) is also a zero, the graph does not change sign there. Furthermore, the function does not change sign between consecutive zeros since the graph of the polynomial is continuous. If the polynomial did change sign, this would mean there is another zero and that the zeros aren't consecutive, a contradiction. These two concepts are the basis for another approach to determining the sign of a polynomial function and solving inequalities called the test point method. If we pick any value of \(x\) in the interval (0,3) between the consecutive zeros \(x=0\) and \(x=3\), we can evaluate the function to determine the sign of \(f\) at that point. For example, at \(x=1\), \(y=f(1)=1^4-1^3-6(1)^2=-5 \) is negative. Since the the polynomial does not change sign between consecutive zeros, the polynomial must be negative on the whole interval (0,3). If we pick any value of \(x\) in the interval (-2,0) between the consecutive zeros \(x=-2\) and \(x=0\) such as \(x=-1\) then evaluate \(y=f(-1)=(-1)^4-(-1)^3-6(-1)^2=-4 \), we can make a similar argument to determine that \(f\) is also negative on the interval (-2,0). If we repeated this process of evaluating the function at a test point in each interval between consecutive zeros, we could determine the sign of the polynomial \(f\) for all values of \(x\). Notice, we do not need the multiplicity of the zero, the end behavior, or the graph itself to determine the sign of the polynomial on each interval.
Example \(\PageIndex{4}\)
Solve the inequality \(x^4-2x^2 > 0\) using the test point method.
Solution
To solve the inequality we need to determine where \(f(x) =x^4-2x^2 \) is positive. First, we need to find the zeros of \(y=f(x)=x^4-2x^2\) by factoring \[ x^4-2x^2 = x^2(x-2)=0 \nonumber\]Then setting each factor to zero \[ x^2=0 \text{ and } x^2 - 2=0 \nonumber\]The first equation yields a zero of \(x=0\). Solving the second equation \[ x^2=2 \nonumber\] which yields the zeros\[ x=\pm \sqrt{2} \approx 1.41 \nonumber\]These zeros break the entire x-axis into four intervals \( (-\infty,-\sqrt{2}) \), \( (-\sqrt{2},0) \), \( (0,\sqrt{2}) \), and \( (\sqrt{2},\infty) \). Choosing a test value in each interval and evaluating the function at each test value in the table below to determine if the function is positive or negative in each interval
| Interval | Test \(x\) in interval | \(f\) (test value) |
|---|---|---|
| \( (-\infty,-\sqrt{2}) \) | -2 | 8 |
| \( (-\sqrt{2},0) \) | -1 | -1 |
| \( (0,\sqrt{2}) \) | 1 | -1 |
| \( (\sqrt{2},\infty) \) | 2 | 8 |
Therefore, \(f\) is positive on \( (-\infty,-\sqrt{2}) \cup (\sqrt{2},\infty) \) and negative on \( (-\sqrt{2},0) \cup (0,\sqrt{2}) \). We can represent the sign on f with a sign chart as shown below.
The solution to the inequality is \( (-\infty,-\sqrt{2}) \cup (\sqrt{2},\infty) \)
Solve the inequality \(x^2 + x < -2\) using the test point method.
Solution
This inequality is equivalent to the inequality \(x^2 + x +2 < 0\). To solve the inequality we need to determine the zeros of \(y=f(x)=x^2 + x +2\). Using the quadratic formula with \(a=1\), \(b=1\) and \(c=2\), we have \[x= \dfrac{-1 \pm \sqrt{1^2-4(1)(2)}}{2(1)}\nonumber \] \[x= \dfrac{-1 \pm \sqrt{-7}}{2} = \dfrac{-1 \pm i \sqrt{7}}{2} = \dfrac{-1}{2} \pm i \dfrac{\sqrt{7}}{2}\nonumber \] So there are no real zeros. Therefore the function never crosses the x-axis and does not change sign. In this case, we need only test one point in the interval \( (-\infty,\infty) \).
| Interval | Test \(x\) in interval | \(f\) (test value) |
|---|---|---|
| \( (-\infty,\infty)) \) | 0 | 2 |
Therefore, \(f\) is positive for all values of x (on \( (-\infty,-\infty) \)). \(f\) is never negative. There is no solution to the inequality.
Solve the inequality \(x^3 + 9x^2 + -20x \ge 0 \) using the test point method. Build a sign chart for \( f(x)=x^3 + 9x^2 +20x \) with your results.
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We need to determine where \( f(x)=x^3 + 9x^2 +20x \) is positive or zero. The zeros of \( f(x)=x^3 + 9x^2 +20x \) are \(x=0\), \(x=-4\), and \(x=-5\). Choosing a test value in each interval and evaluating the function at each test value in the table below
Test Points and Their Values in Each Interval Interval Test \(x\) in interval \(f\) (test value) \( (-\infty,-5) \) -6 -12 \( (-5,-4) \) -4.5 1.125 \( (-4,0) \) -1 -12 \( (0,\infty) \) 1 30 Therefore, \(f\) is positive or zero on \( [-5,-4] \cup [0,\infty) \). These intervals are the solutions to the original inequality.
The sign chart for \(f\) is given below.
Important Topics of this Section
- Solving Polynomial Inequalities
- The Test Point Method