Problem 43 - College Algebra with Modeling and Visualization

Question

Solve the polynomial inequality (a) symbolically and (b) graphically. $$ x^{3}+x^{2} \geq 2 x $$

Step-by-Step Solution

Answer

The solution is \((-2, 0] \cup [1, \infty)\).

Step 1: Rewrite the Inequality

To solve the inequality \(x^3 + x^2 \geq 2x\), start by moving all terms to one side to get a single inequality: \(x^3 + x^2 - 2x \geq 0\).

Step 2: Factor the Polynomial

Factor the polynomial \(x^3 + x^2 - 2x\). Start by factoring out the common term \(x\): \(x(x^2 + x - 2) \geq 0\). Then, factor the quadratic \(x^2 + x - 2\) to get \((x - 1)(x + 2)\). So, the inequality becomes \(x(x - 1)(x + 2) \geq 0\).

Step 3: Identify Critical Points

Set each factor equal to zero to find the critical points: \(x = 0\), \(x = 1\), and \(x = -2\). These points divide the number line into intervals.

Step 4: Test Intervals

Test each interval determined by the critical points to see where the product is non-negative. The intervals are: \((-\infty, -2)\), \((-2, 0)\), \((0, 1)\), and \((1, \infty)\). Choose a test point from each interval and substitute it into \(x(x - 1)(x + 2)\) to determine the sign: - For \((-\infty, -2)\), test \(x = -3\): negative.- For \((-2, 0)\), test \(x = -1\): positive.- For \((0, 1)\), test \(x = 0.5\): negative.- For \((1, \infty)\), test \(x = 2\): positive.

Step 5: Write the Solution Symbolically

The solution is where the product is non-negative. This occurs in the intervals \((-2, 0] \cup [1, \infty)\), including \(x = 0\) and \(x = 1\), where the product is zero.

Step 6: Solve Graphically

To solve graphically, sketch the graph of \(y = x^3 + x^2 - 2x\). The graph crosses the x-axis at \(x = -2\), \(0\), and \(1\). Observe where the graph is above (or on) the x-axis, which corresponds to the solution found symbolically. It is above the x-axis in the intervals \((-2, 0] \cup [1, \infty)\).

Step 7: Verify the Solution

Confirm the solution by checking the critical points and the sign change at each interval. Both the symbolic and graphical solutions should match the intervals: \((-2, 0] \cup [1, \infty)\).

Key Concepts

Factoring Polynomials

Factoring polynomials is a key step in solving polynomial inequalities. It involves breaking down a polynomial into simpler "factor" polynomials that, when multiplied together, produce the original polynomial. In our exercise, we started with the inequality \(x^3 + x^2 - 2x \geq 0\).

First, we factored out the common factor \(x\) from each term:

• The expression \(x(x^2 + x - 2)\) was formed.

Next, we factorized the quadratic \(x^2 + x - 2\) into

• \((x - 1)(x + 2)\).

These factored terms allowed us to rewrite the polynomial as \(x(x - 1)(x + 2) \geq 0\). This simplification is crucial as it prepares the inequality for identifying critical points and subsequently testing intervals.

Critical Points

Critical points in a polynomial inequality are values of \(x\) where the polynomial equals zero. These points are essential because they mark transitions on the graph, potentially indicating changes in inequality sign.

For \(x(x - 1)(x + 2) \geq 0\), we determined the critical points by setting each factor equal to zero:

• \(x = 0\)
• \(x = 1\)
• \(x = -2\)

These critical points divide the number line into distinct intervals where we can test the polynomial's sign. Remember, finding critical points is pivotal as they help identify where the inequality holds true or changes direction.

Graphical Solution

Visualizing the polynomial can be an intuitive way to solve inequalities. When graphing \(y = x^3 + x^2 - 2x\), you're observing the curve's interaction with the x-axis, where the value of \(y = 0\).

In the graphical solution, the polynomial crosses the x-axis at the critical points:

• \(x = -2\)
• \(x = 0\)
• \(x = 1\)

The solution to the inequality is where the graph is above or touches the x-axis (\((y \geq 0)\)). Observing the graph, you see it remains above or on the x-axis within the intervals

• \((-2, 0]\)
• \([1, \infty)\)

The graphical method complements symbolic solutions and is especially useful for confirming results.

Interval Testing

Interval testing involves choosing sample points from different intervals, determined by the critical points, to decide where the inequality holds. For the polynomial \(x(x - 1)(x + 2)\), we had these intervals:

• \((-\infty, -2)\)
• \((-2, 0)\)
• \((0, 1)\)
• \((1, \infty)\)

By picking a test point from each interval and substituting it back into the factors:

• \((-\infty, -2): \; x = -3\) (negative)
• \((-2, 0): \; x = -1\) (positive)
• \((0, 1): \; x = 0.5\) (negative)
• \((1, \infty): \; x = 2\) (positive)

These tests allowed us to detect where the polynomial's product was non-negative. It's a crucial step to establish the specific regions where the inequality holds true, especially when complemented by the critical points.