PreCalculus

7.3. Solving polynomial inequalities

A polynomial inequality can be written as a product of linear factors on one side of an inequality symbol and with zero on the other side; for example:

(x + 1)(x - 2) < 0

This can be solved by use of a sign diagram. The point where a factor is equal to zero is called a critical point. For the polynomial above, the critical points are -1, 2. The value of the polynomial changes sign only at the critical points. For a quadratic with real zeroes, the number line can be divided into three regions where the value of the quadratic changes sign. If the inequality is true for one point within a particular region then it is true for all points in that region. Similarly, if the inequality is false for some point in a particular region then it is false for every point in that region.

Factoring a polynomial allows us to create a map that shows where each factor is negative or positive. By using the rules for multiplying signed quantities, we can find the sign of the value of the polynomial in each region. So, to solve a polynomial inequality, we perform the following steps:

Rewrite the inequality so that there is a zero on the right side.
Find all linear factors of the function.
To find the critical points, set each linear factor to zero and solve for x.
Determine the sign of the polynomial in the intervals formed by the critical points.
The solution will be those intervals in which the polynomial has the sign that satisfies the inequality.

Example, solve: x² - x < 2

x² - x - 2 < 0
(x - 2)(x + 1) < 0
The critical points are x = -1, 2

(x - 2)     -  |        -        |  +
(x + 1)     -  |        +        |  +
Product     +  |        -        |  +
        -------+-----------------+------
              -1     0     1     2

Then, the inequality is true in the open interval: (-1, 2)

Example, solve: x² - 3 ≥ 2x

x² - 2x - 3 ≥ 0 
(x + 1)(x - 3) ≥ 0, the critical points are x = -1, x = 3

(x - 3)     -  |           -           |  +
(x + 1)     -  |           +           |  +
Product     +  |           -           |  +
        -------+-----------------------+------
              -1     0     1     2     3

Then, the inequality is true in the two half open intervals: (-∞, -1], [3, +∞)

Example, solve: x³ + 2x² - 5x - 6 > 0

(x + 1)(x - 2)(x + 3) > 0
Then, the critical points are: x = -3 x = -1 x = 2, 
These divide the number line into four intervals.

(x + 3)     -  |     +     |        +        |  +
(x + 1)     -  |     -     |        +        |  +
(x - 2)     -  |     -     |        -        |  +
Product     -  |     +     |        -        |  +
        -------+-----------+-----------------+-----
              -3    -2    -1     0     1     2

Then, the inequality is true in the two intervals: (-3, -1), (2, +∞)