Problem 13 - Precalculus : Building Concepts and Connections

Question

Solve the polynomial inequality. $$x^{3}+5 x^{2}+4 x<0$$

Step-by-Step Solution

Answer

The solution of the inequality is -4 < x < -1 and -1 < x < 0.

Step 1: Simplify the inequality

Looking at the given inequality, it can be seen that there is a common factor of x that can be factored out. This will simplify the inequality. After factoring out, the inequality becomes \(x*(x^{2}+5x+4)<0\).

Step 2: Factoring the quadratic expression

The quadratic \(x^{2}+5 x+4\) in the inequality can be factored further. After factoring, the inequality becomes \(x*(x+1)(x+4)<0\).

Step 3: Identify the critical points

The next step is to find the critical points of the inequality, which lie where the expression equals zero. Thus the critical points are at x=-4, x=-1, and x=0.

Step 4: Test values to check inequality

Now choose values within the intervals between critical values and plug into the expression. For x<-4 try x=-5, for -40 try x=1. It is found that the inequality holds true at the intervals (-4,-1) and (-1,0).

Step 5: Present the solution of the inequality

So, the solution of the given inequality \(x^{3}+5 x^{2}+4 x<0\) is -4 < x < -1 and -1 < x < 0.

Key Concepts

Factoring Polynomials

Factoring is a key step in solving polynomial inequalities because it breaks down complex expressions into simpler, more manageable parts. In our example, the polynomial inequality x^{3}+5x^{2}+4x<0 is approached by initially spotting a common factor among all terms, which is x. Factoring out x simplifies the inequality to x(x^{2}+5x+4)<0.

The remaining quadratic x^{2}+5x+4 is then factored further, which involves finding two numbers that multiply to the constant term (in this case, 4) and add up to the linear coefficient (in this case, 5). These numbers are 1 and 4, hence we factor the quadratic to (x+1)(x+4). The original inequality now reads x(x+1)(x+4)<0, where each factor can be analyzed separately to find the solution set for the inequality. Factoring polynomials is crucial as it allows us to identify the critical points of the inequality, setting us up for the next steps in the solution process.

Critical Points in Inequalities

Critical points are values that make the polynomial equal to zero and are essential in determining the intervals to test for the solution of an inequality. In the context of our solved inequality, once we have factored the polynomial as x(x+1)(x+4), we set each factor to zero to find the critical points. This yields the critical values x=0, x=-1, and x=-4.

These points divide the number line into intervals that we can test to determine where the original inequality holds true. The critical values themselves are not included in the solution set for a strict inequality (< or >), but they are for a non-strict inequality (≤ or ≥). Inequalities involve understanding the behavior of the polynomial expression across these intervals which is key to solving them. By knowing exactly where the expression changes sign, we can accurately define where the inequality is satisfied.

Test Values in Inequalities

Once we have established critical points, the next step involves selecting test values. These values are chosen from each of the intervals created by the critical points. The goal is to determine which intervals satisfy the original inequality. In our exercise, test values such as x=-5, x=-3, x=-0.5, and x=1 are used to evaluate the inequality within the intervals (-∞, -4), (-4, -1), (-1, 0), and (0, +∞), respectively.

Inserting these test values into the factored inequality helps us find that the inequality x^{3}+5x^{2}+4x<0 holds true for -4 < x < -1 and -1 < x < 0. Choosing the right test values is crucial: they must be easy to compute and clearly indicate the sign of the expression within the interval. Hence, the intervals where the expression is negative are part of our solution to the inequality, providing a clear understanding of where the original inequality is valid.