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Chapter 8: Problem 25

Quadratic and Other Polynomial Inequalities Solve. $$(x-1)(x+2)(x-4) \geq 0$$

Short Answer

Expert verified
The solution is \[(-\infty, -2] \cup [-2, 1] \cup [4, \infty)\].

Step by step solution

01

Determine the Critical Points

Solve the equation \((x-1)(x+2)(x-4) = 0\) to find the critical points. The solutions are: \(x = 1\), \(x = -2\), and \(x = 4\). These values are where the expression could change sign.
02

Determine the Intervals

Using the critical points, the number line is divided into the following intervals: \((-\infty, -2)\), \((-2, 1)\), \((1, 4)\), and \((4, \infty)\).
03

Test Each Interval

Select a test point within each interval and substitute it into the inequality \((x-1)(x+2)(x-4)\) to determine if the inequality holds: For \((-\infty, -2)\), use \(x = -3\): \[(-3-1)(-3+2)(-3-4) = (-4)(-1)(-7) > 0\] (Positive) For \((-2, 1)\), use \(x = 0\): \[(0-1)(0+2)(0-4) = (-1)(2)(-4) > 0\] (Positive) For \((1, 4)\), use \(x = 2\): \[(2-1)(2+2)(2-4) = (1)(4)(-2) < 0\] (Negative) For \((4, \infty)\), use \(x = 5\): \[(5-1)(5+2)(5-4) = (4)(7)(1) > 0\] (Positive)
04

Include Critical Points

Since the inequality \((x-1)(x+2)(x-4) \geq 0\) includes equality, the critical points \(x = -2\), \(x = 1\), and \(x = 4\) should be included in the solution.
05

Write the Solution Set

From the testing of intervals, the solution set includes all the intervals where the expression is non-negative: \[(-\infty, -2] \cup [-2, 1] \cup [4, \infty)\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Inequalities
Understanding quadratic inequalities is key to solving polynomial inequalities like \((x-1)(x+2)(x-4) \geq 0\). Quadratic inequalities involve expressions like \[ax^2 + bx + c = 0.\] The solutions determine the ranges of x that satisfy the inequality. Working through these inequalities, you can visualize these ranges often using a number line, testing different values to see where the inequality holds.
Critical Points
The critical points are the values where the inequality changes sign. To find them, solve the equation by setting the polynomial equal to zero, as in \((x-1)(x+2)(x-4) = 0\). Solving this gives you the points \{x = 1, x = -2, x = 4\}. These points are significant because they help you divide the number line into intervals to test the polynomial's behavior on those intervals.
Interval Testing
Interval testing helps determine whether the polynomial is positive or negative in those intervals. After finding critical points, create intervals between and around those points: \((-\infty, -2)\), \((-2, 1)\), \((1, 4)\), and \((4, \infty)\). Next, pick a test point from each interval, substitute these into the polynomial and observe the results. For instance, using \[x = 0\] in \((-2, 1)\) results in a positive expression, therefore, that interval satisfies the inequality.
Solution Sets
The solution set includes all intervals where the polynomial is non-negative. After interval testing, gather the results where the value is positive or zero. This example includes the intervals \((-\infty, -2] \cup [-2, 1] \cup [4, \infty)\). Here, incorporate the critical points into the solution since \geq\ implies equality. The final solution set is a combination of all these segments.

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