Problem 38 - Intermediate Algebra

Question

Solve each inequality. $$ 2 x^{3}+4 x^{2} \leq 0 $$

Step-by-Step Solution

Answer

The solution to the inequality is \([-2, 0]\).

Step 1: Identify Mathematical Expression

The given inequality is \(2x^3 + 4x^2 \leq 0\). We need to solve for the range of values of \(x\) that satisfy this inequality.

Step 2: Factor the Inequality

Factor the expression on the left side of the inequality:\[2x^3 + 4x^2 = 2x^2(x + 2)\]

Step 3: Solve the Corresponding Equation

Set the factored expression equal to zero to find the critical points:\[2x^2(x + 2) = 0\]This gives the critical points:1. \(2x^2 = 0\) which simplifies to \(x = 0\).2. \(x + 2 = 0\) which simplifies to \(x = -2\).

Step 4: Analyze the Sign of Each Interval

Using \(x = 0\) and \(x = -2\) as critical points, set up intervals:1. \((-\infty, -2)\)2. \((-2, 0)\)3. \((0, \infty)\)Test a point from each interval to determine where the expression \(2x^2(x + 2)\) is negative or zero.

Step 5: Test Intervals

Test a point in each interval:- For \(x = -3\) in \((-\infty, -2)\), the expression is \(2(-3)^2(-3 + 2) = 18(-1) = -18\) which is \(\leq 0\).- For \(x = -1\) in \((-2, 0)\), the expression is \(2(-1)^2(-1 + 2) = 2\) which is \(\geq 0\).- For \(x = 1\) in \((0, \infty)\), the expression is \(2(1)^2(1 + 2) = 6\) which is \(\geq 0\).

Step 6: Construct the Solution

The expression is non-positive (\(\leq 0\)) in the interval where the test value is negative or zero. From our test, this occurs in the interval \([-2, 0]\). Both \(x = -2\) and \(x = 0\) satisfy the equation, so they are included.

Key Concepts

Polynomial Factorization

To solve polynomial inequalities, polynomial factorization is a crucial step. In our exercise, we started with the expression \(2x^3 + 4x^2 \leq 0\). Factorization means expressing this polynomial as a product of simpler expressions. The key is to find common factors first. Here, the term \(2x^2\) appeared in both parts of the polynomial.
We factor it out like this:

• Extract the common factor: \(2x^2(x + 2)\)

By breaking down the expression into products, we can explore values of \(x\) that make it equal to zero. These are our critical points. Factorization makes it easier to visualize how the equality behaves across different ranges of \(x\). Understanding this factorization lays the groundwork for finding where the entire polynomial is less than or equal to zero.

Critical Points

Finding critical points is like finding the puzzle pieces to the inequality. Once we have factored our polynomial into \(2x^2(x + 2)\), we set each part of this product to zero to find points where the polynomial changes its behavior.

• \(2x^2 = 0\) implies \(x = 0\).
• \(x + 2 = 0\) implies \(x = -2\).

These solutions \(x = 0\) and \(x = -2\) are critical because they divide the number line into different sections where the polynomial will either be positive or negative. Critical points help us determine the intervals to test and understand the behavior of the polynomial across these segments.

Interval Testing

Interval testing helps us understand which sections of the number line make the inequality \(2x^3 + 4x^2 \leq 0\) true. We've identified the critical points as \(x = 0\) and \(x = -2\). These points divide the number line into intervals:

• \((−\infty, −2)\)
• \((−2, 0)\)
• \((0, \infty)\)

We then pick a test point from each interval:

• For \(x = -3\) in \((−\infty, −2)\), the polynomial is negative.
• For \(x = -1\) in \((−2, 0)\), the polynomial is positive.
• For \(x = 1\) in \((0, \infty)\), the polynomial is positive.

Testing helps determine where the inequality \(\leq 0\) holds true. This comprehensively covers the polynomial behavior across its domain. Understanding interval testing is an essential method to solve and verify solutions for inequalities by testing sample values and observing resultant signs.