Problem 10 - Precalculus : Building Concepts and Connections

Question

Solve the polynomial inequality. $$x^{3}-9 x>0$$

Step-by-Step Solution

Answer

The solution of the inequality is \(x ∈ (-∞, -3) \cup (0, 3)\).

Step 1: Factor the inequality

To solve polynomial inequalities, first factorize the polynomial. The inequality \(x^{3} - 9x > 0\) can be factored to \(x(x^{2} - 9) > 0\). This can further be factored into \(x(x - 3)(x + 3) > 0\) using difference of squares.

Step 2: Find the critical points

The critical points are the roots of the equation \(x(x - 3)(x + 3) = 0\). Set each factor equal to zero and solve for \(x\). There are three solutions: \(x = 0, x = 3, x = -3\). These points split the real number line into intervals.

Step 3: Test the intervals

Test each interval in the inequality, using a test point. The intervals to test are \((-∞, -3), (-3, 0), (0, 3), (3, ∞)\). If a test point makes the inequality true, then the entire interval is in the solution set. Take -4 for the first interval, -1 for the second interval, 2 for the third interval and 4 for the fourth interval. After plugging these test points into the inequality, the true intervals are \((-∞, -3)\) and \((0, 3)\).

Step 4: Write the final solution

The solution of the inequality is the union of the intervals where the inequality holds true. Thus, the solution is \(x ∈ (-∞, -3) \cup (0, 3)\). Since the inequality sign does not include equal, the critical points are not part of the solution set and thus the solution to the inequality is written with parentheses, not brackets.

Key Concepts

Factoring Polynomials

Factoring polynomials is a crucial step in solving polynomial inequalities. Let's take the inequality given to us:

\[ x^3 - 9x > 0 \]
To simplify this problem, we need to express the polynomial as a product of its factors. A factored form often reveals the roots or critical points of the polynomial, making it easier to analyze.

For this particular case, we can first factor out an \(x\), yielding:

• \(x(x^2 - 9) > 0\)

Next, notice that \(x^2 - 9\) is a difference of squares, which can be further stratified:

• \(x(x - 3)(x + 3) > 0\)

Factoring in this way reduces the complexity of the polynomial and sets the stage for the next steps in solving the inequality.

Critical Points

Critical points are where the polynomial equals zero, dividing the number line into intervals we can analyze. These points arise naturally from the factors of the polynomial.

In the factored inequality:

• \(x(x - 3)(x + 3) > 0\)

Setting each factor to zero gives us the critical points:

• \(x = 0\)
• \(x = 3\)
• \(x = -3\)

These zeros or roots separate the real number line into distinct intervals. Identifying critical points is fundamental, as it highlights where changes in the sign of the polynomial can occur. Throughout these intervals, you will find that the polynomial is either entirely positive or negative, unless you are precisely at a critical point.

Interval Testing

Interval testing involves choosing test points from each interval formed by the critical points and checking them against the inequality. This tells us whether the inequality holds true in those ranges.

For our polynomial

• \(x(x - 3)(x + 3) > 0\)

the critical points

• -3, 0, and 3

create four testable intervals:

• \((-∞, -3)\)
• \((-3, 0)\)
• \((0, 3)\)
• \((3, ∞)\)

Selecting values within these intervals, such as

• -4 for \((-∞, -3)\)
• -1 for \((-3, 0)\)
• 2 for \((0, 3)\)
• 4 for \((3, ∞)\)

we test each point by substituting back into the inequality. Through this process, identify which intervals render the inequality true, thus indicating where the polynomial is greater than zero.

Solution Sets

The solution set for a polynomial inequality comprises all the intervals where the inequality conditions are satisfied. After testing, these intervals form our solution. For the inequality

• \(x(x - 3)(x + 3) > 0\)

the interval testing showed us that the polynomial is positive in:

• \((-∞, -3)\)
• \((0, 3)\)

Therefore, the solution set can be written as the union of these intervals: \[ x ∈ (-∞, -3) \, ∪ \, (0, 3) \]It's important to note that the critical points are not part of the solution set since they result in the polynomial value being zero, not greater than zero. Thus, we use parentheses to denote intervals, ensuring clarity that the endpoints aren't included in the solutions.