Problem 35 - Algebra and Trigonometry

Question

Solve each polynomial inequality in Exercises \(1-42\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ x^{3}+2 x^{2}-x-2 \geq 0 $$

Step-by-Step Solution

Answer

The solution to the inequality \(x^{3}+2x^{2}-x-2\geq 0\) is \((-2,-1] \cup (1, \infty)\).

Step 1: Solve the Polynomial

First, the roots of the polynomial \(x^{3}+2x^{2}-x-2\) are found by factoring it. One possible factorization gives \((x - 1)(x + 1)(x + 2) = 0\). Setting each factor equal to zero gives the roots as \(x = -2, -1, 1\).

Step 2: Interval Test Values

Determine some test values for the intervals formed by these roots. Split the real number line into four intervals according to the roots: \(-\infty \lt x \lt -2\), \(-2 \lt x \lt -1\), \(-1 \lt x \lt 1\) and \(1 \lt x \lt \infty\). Choose \(-3\) for the first interval, \(-1.5\) for the second interval, \(0\) the third interval, and \(2\) for the fourth interval.

Step 3: Check the Inequality Conditions

Check whether the test values satisfy the inequality \(x^{3}+2x^{2}-x-2\geq 0\). For \(-3\), it yields \(-\), hence does not satisfy the condition. \(-1.5\) yields \(+\), hence it satisfies the condition. \(0\) yields \(-\), hence it doesn't satisfy the condition. \(2\) yields \(+\), hence it satisfies the condition.

Step 4: Write the final answer in interval notation

Since the values from the second and fourth intervals, \(-2 \lt x \lt -1\) and \(1 \lt x \lt \infty\), satisfied the inequality, the solution set in interval notation is \((-2,-1] \cup (1, \infty)\).