▪ Solving Polynomial Inequalities | College Algebra Corequisite

Learning Outcomes

Solve polynomial inequalities using boundary value method.

Solving Polynomial Inequalities using Boundary Value Method

Any inequality that can be put into one of the following forms

$f(x)>0, f(x) \geq 0, f(x)<0,$ or $f(x) \leq 0$ , where $f$ is a polynomial function

is called polynomial inequality.

How To: Solve Polynomial Inequalities using Boundary Value MEthod

Rewrite the given polynomial inequality as an equation by replacing the inequality symbol with the equal sign.
Solve the polynomial equation. The real solution(s) of the equation is(are) the boundary point(s).
Plot the boundary point(s) from Step 2 on a number line.
$\Rightarrow$ Use an open circle when the given inequality has $<$ or $>$
$\Rightarrow$ Use a closed circle when the given inequality has $\leq$ or $\geq$ .
Choose one number, which is called a test value, from each interval and test the intervals by evaluating the given inequality at that number.
$\Rightarrow$ If the inequality is TRUE, then the interval is a solution of the inequality.
$\Rightarrow$ If the inequality is FALSE, then the interval is not a solution of the inequality.
Write the solution set (usually in interval notation), selecting the interval(s) from Step 4.

Example: Solving Polynomial Inequality using Boundary Value Method

Solve the polynomial inequality using boundary value method. Graph the solution set and write the solution in interval notation.

$x^4-4x^3+3x^2 > 0$

Show Solution

First, let’s rewrite the inequality as an equation and then solve it:

$x^4-4x^3+3x^2=0$

$x^2(x^2-4x+3)=0$

$x^2(x-1)(x-3)=0$

$x^2=0, x-1=0$ or $x-3=0$

$x=0, x=1$ or $x=3$

Second, plot the solutions on the number line. Since the inequality has “ $>$ ” (greater than), which doesn’t have “ $=$ ” (equal sign), they should be open circles:

Numberline plotted 0, 1, 3 (open) as solutions, marked -1, 0.5, 2, 5 as testing values

Note that now there are four intervals on the number line: $(-\infty, 0), (0, 1), (1, 3),$ and $(3, \infty)$

Third, choose your choice of test value from each interval:

I chose $-1, 0.5, 2,$ and $5$ from those four intervals.

Fourth, test each interval using the test value:

When $x=-1$ , $(-1)^4-4(-1)^3+3(-1)^2>0 \rightarrow 8>0$ $\therefore$ solution

When $x=0.5$ , $(0.5)^4-4(0.5)^3+3(0.5)^2>0 \rightarrow 0.3125>0$ $\therefore$ solution

When $x=2$ , $(2)^4-4(2)^3+3(2)^2>0 \rightarrow -4>0$ $\therefore$ not a solution

When $x=5$ , $(5)^4-4(5)^3+3(5)^2>0 \rightarrow 200>0$ $\therefore$ solution

Number line plotted 0, 1, 3 (open) as solutions, marked -1, 0.5, 2, 5 as testing values, first and fourth intervals are marked in red

Actually, we can check the results of test values from the graph of $f(x)=x^4-4x^3+3x^2$ below:

Graph of f(x)=x^4-4x^3+3x^2, marked above x-axis in red

* As we can see from the graph, it doesn’t matter which number you choose from each interval for testing. For example, any number that is greater than 5 will make the function value positive.

Therefore, the solution of the polynomial inequality $x^4-4x^3+3x^2 > 0$ is $(-\infty, 0) \cup (0, 1) \cup (3, \infty)$ .

Try It

Solve the polynomial inequality using boundary value method. Graph the solution set and write the solution in interval notation.

$x^2 \leq 3x+4$

Show Solution

First, let’s rewrite the inequality as an equation and then solve it:

$x^2 = 3x+4$

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

$(x+1)(x-4)=0$

$x+1=0$ or $x-4=0$

$x=-1$ or $x=4$

Second, plot the solutions on the number line. Since the inequality has “ $\leq$ ” (less than or equal to), which has “ $=$ ” (equal sign), they should be closed circles:

Number line plotted -1, 4 (closed) as solutions, marked -2, 0, 5 as test values

Note that now there are three intervals on the number line: $(-\infty, -1], [-1, 4],$ and $[4, \infty)$

Third, choose your choice of test value from each interval:

I chose $-2, 0,$ and $5$ from those three intervals.

Fourth, test each interval using the test value:

When $x=-2$ , $(-2)^2 \leq 3(-2)+4 \rightarrow 4 \leq -2$ $\therefore$ not a solution

When $x=0$ , $(0)^2 \leq 3(0)+4 \rightarrow 0 \leq 4$ $\therefore$ solution

When $x=5$ , $(5)^2 \leq 3(5)+4 \rightarrow 25 \leq 19$ $\therefore$ not a solution

Number line plotted -1, 4 (closed) as solutions, marked -2, 0, 5 as test values, middle interval is marked in red

We can see the results of test values are true from the graph of $f(x)=x^2-3x-4$ below:

Graph of f(x)=x^2-3x-4, marked below x-axis in red

Therefore, the solution of the polynomial inequality $x^2 \leq 3x+4$ is $[-1, 4]$ .

Example: No solution Case

Solve the polynomial inequality using boundary value method. Graph the solution set and write the solution in interval notation.

$x^6 +2 < 0$

Show Solution

First, let’s rewrite the inequality as an equation and then solve it:

$x^6 + 2 = 0$

$x^6 \ne -2$

So, there is no solution for this polynomial equation.

Second, since there is no solution, we cannot plot anything on the number line, and there is only one interval on it: [latex](-\infty, \infty) Number line no solution plotted, marked 0 as test value Third, choose your choice of test value from the interval:

I chose $0$ .

Fourth, test the interval using the test value:

When $x=0$ , $(0)^6+2<0 \rightarrow 2<0$ $\therefore$ not a solution

Number line no solution plotted, marked 0 as test value

We can see the result of test value is true from the graph of $f(x)=x^2-3x-4$ below. The graph is above the $x$ -axis everywhere:

Graph of f(x)=x^6+2, marked below x-axis in red

Therefore, the solution of the polynomial inequality $x^6+2>0$ is no solution.

Try It

Solve the polynomial inequality using boundary value method. Graph the solution set and write the solution in interval notation.

$-\frac{1}{3}x^2 \leq 0$

Show Solution

First, let's rewrite the inequality as an equation and then solve it:

$-\frac{1}{3}x^2 = 0$

$x^2=0$

$x=0$

Second, plot the solution on the number line. Since the inequality has " $\geq$ " (greater than or equal to), which has " $=$ " (equal sign), it should be closed circles:

Number line plotted 0 (closed) as solution, marked -1, 1 as test values,

Third, choose your choice of test value from the interval:

I chose $-1$ and $1$ .

Fourth, test the interval using the test value:

When $x=-1$ , $-\frac{1}{3}(-1)^2 \leq 0 \rightarrow -\frac{1}{3} \leq 0$ $\therefore$ solution

When $x=1$ , $-\frac{1}{3}(1)^2 \leq 0 \rightarrow -\frac{1}{3} \leq 0$ $\therefore$ solution

Number line plotted 0 (closed) as solution, marked -1, 1 as test values, whole number line is marked in red

We can see the result of test values are true from the graph of $f(x)=-\frac{1}{3} x^2$ below:

Graph of f(x)=-1/3x^2, marked below x-axis in red

* The graph is below the $x$ -axis everywhere except at $x=0$ . However, since the inequality has " $\leq$ ," which has " $=$ " (equal sign), $x=0$ also satisfies the inequality.

Therefore, the solution of the polynomial inequality $-\frac{1}{3} x^2$ is $(-\infty, \infty)$ .

Module 1: Equations and Inequalities