The Law of Trichotomy – Polynomials in General Part I

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My primary recommendation is that we always simultaneously consider an equation and its two inequality siblings as a group and that we graph all three on the same number line.

Law of Trichotomy Applied to Polynomial Inequalities in One Variable:

Rule: The procedures demonstrated in Example 1 and Example 2 (of previous two posts) may be generalized to devise a method to solve any polynomial inequality in one variable.

This will be illustrated in Example 3.
Example 3: Solve the inequality x(x – 2)(x + 3)(x + 3)(x + 4) < 0.
Discussion and Solution:
Generally, fifth degree polynomial inequalities in one variable will not be solvable, but when we can factor it, as in this case, we can solve the inequality

When considering the inequality x(x – 2)(x + 3)(x + 3)(x + 4) < 0 the Law of Trichotomy dictates that we be aware of the equality x(x – 2)(x + 3)(x + 3)(x + 4) = 0 as well as the inequality x(x – 2)(x + 3)(x + 3)(x + 4) > 0.The equation x(x – 2)(x + 3)(x + 3)(x + 4) = 0 is called the boundary equation because its graph is the boundary between the graphs of the two inequalities.

Our strategy will be to graph the equation and test numbers from the various resulting rays and intervals formed by that graph. The Zero Factor Property shows the solution set of the equation x(x – 2)(x + 3)(x + 3)(x + 4) = 0 to be {– 4, –3, 0, 2}. A sketch of the graph of the equation x(x – 2)(x + 3)(x + 3)(x + 4) = 0 is shown in Fig. 6. Observe the rays and intervals are;

(– ∞, – 4), (–4, –3), (–3, 0), (0, 2), and (2, ∞)

The solution set for the two inequalities can be determined by testing a number from each of the rays and intervals. However, there is a simpler approach which depends on the fact that it is simple to determine the sign of a linear expression in an interval.

A convenient and effective way to organize the test is with a table as illustrated in Fig. 7.

The final row of this table shows where the polynomial is positive and where it is negative. We therefore conclude that:
The solution set for x(x – 2)(x + 3)(x + 3)(x + 4) = 0 is
{–4, –3, 0, 2}.
The solution set for x(x – 2)(x + 3)(x + 3)(x + 4) > 0 is
(–4, –3) ∪ (–3, 0) ∪ (2, ∞).
The solution set for x(x – 2)(x + 3)(x + 3)(x + 4) < 0 is
(–∞, –4) ∪ (0, 2).

A Final Note:  I have been using the term “solution set” rather than “solution”.  Recall the definitions:
A solution is a number or numbers which produces a true statement when substituted for the variable(s).
A solution set for an equation or inequality is the set of all solutions of the equation or inequality.

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