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Chapter 4: Problem 61

Solve each rational inequality by hand. Do not use a calculator. $$\frac{(x+1)(x-2)}{(x+3)}<0$$

Short Answer

Expert verified
The solution is \((-\infty, -3) \cup (-1, 2)\).

Step by step solution

01

Identifying the Problem Structure

The given inequality is \( \frac{(x+1)(x-2)}{(x+3)}<0 \). This is a rational inequality since it involves a fraction and a polynomial. Our goal is to find the values of \( x \) that make this expression negative.
02

Finding the Critical Points

The critical points are found by setting each factor in the numerator and the denominator equal to zero: 1. \( x + 1 = 0 \) gives \( x = -1 \).2. \( x - 2 = 0 \) gives \( x = 2 \).3. \( x + 3 = 0 \) gives \( x = -3 \).These points are important as they are potential locations where the inequality changes its sign.
03

Creating the Test Intervals

The critical points divide the number line into different intervals: \(( -\infty, -3 )\), \((-3, -1)\), \((-1, 2)\), and \((2, \infty)\). We need to test each interval to see where the inequality holds.
04

Testing the Inequality on the Intervals

Choose a test point from each interval to determine the sign of the expression:1. **Interval \((-\infty, -3)\):** Test \( x = -4 \). - \( \frac{((-4)+1)((-4)-2)}{(-4+3)} = \frac{(-3)(-6)}{-1} = \frac{18}{-1} = -18 < 0 \), inequality holds.2. **Interval \((-3, -1)\):** Test \( x = -2 \). - \( \frac{((-2)+1)((-2)-2)}{(-2+3)} = \frac{(-1)(-4)}{1} = \frac{4}{1} = 4 > 0 \), inequality does not hold.3. **Interval \((-1, 2)\):** Test \( x = 0 \). - \( \frac{(0+1)(0-2)}{(0+3)} = \frac{(1)(-2)}{3} = \frac{-2}{3} < 0 \), inequality holds.4. **Interval \((2, \infty)\):** Test \( x = 3 \). - \( \frac{(3+1)(3-2)}{(3+3)} = \frac{(4)(1)}{6} = \frac{4}{6} = \frac{2}{3} > 0 \), inequality does not hold.
05

Considering the Critical Points

The inequality \(\frac{(x+1)(x-2)}{(x+3)} < 0\) does not include the points where the expression is undefined or zero, so we exclude \(x = -3\), \(x = -1\), and \(x = 2\) from the solution.
06

Writing the Solution

Based on our analysis, the solution to the inequality is the union of the intervals where the inequality holds: \((-\infty, -3)\) and \((-1, 2)\).

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Key Concepts

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

Polynomial Factorization
Polynomial factorization is a key step when solving rational inequalities. It involves breaking down a polynomial into simpler, non-divisible polynomials, known as factors. In our example, the numerator of the inequality \( \frac{(x+1)(x-2)}{(x+3)}<0 \) is already factored into \( (x+1) \) and \( (x-2) \). This simplifies the process of finding critical points, as it allows us to easily identify the values of \( x \) that make these factors equal to zero.
  • Why Factorize? Factorizing helps us understand how the expression behaves, especially how it changes sign around its critical points.
  • Steps for Factorization: Identify patterns like the difference of squares, common factors, or trinomial structures that can be split into binomials.
Start with identifying each polynomial in the rational expression. Then work to express them in their simplest factor form, making critical point calculation straightforward.
Critical Points
Finding critical points is essential in determining where an expression potentially changes its sign. Critical points are found by setting the numerator and denominator of the factored form of the inequality to zero. For the inequality \( \frac{(x+1)(x-2)}{(x+3)}<0 \), the critical points are:
  • From \( x+1 = 0 \), we find \( x = -1 \).
  • From \( x-2 = 0 \), we find \( x = 2 \).
  • From \( x+3 = 0 \), we find \( x = -3 \).
These points divide the number line into intervals. Critical points inform us of potential sections on the number line where the expression might switch its sign from positive to negative or vice versa.
Interval Testing
After identifying critical points, we use them to create intervals along the number line. Interval testing involves selecting test points from each interval to determine whether the inequality holds for each section. Our intervals for \( \frac{(x+1)(x-2)}{(x+3)} < 0 \) are:
  • \( (-\infty, -3) \)
  • \( (-3, -1) \)
  • \( (-1, 2) \)
  • \( (2, \infty) \)
Pick a test point from each interval, such as \( x = -4 \) for \( (-\infty, -3) \), and substitute it back into the inequality to check the sign. This process helps identify which intervals make the inequality true (i.e., result in a negative expression). Always remember to omit the critical points themselves in these evaluations unless specified otherwise in the inequality.
Algebraic Solutions
Combining factorization, critical points, and interval testing eventually leads to the algebraic solution of the inequality. The solution is the union of intervals that satisfy the inequality condition. For our problem, the critical testing concluded that the inequality holds for the intervals:
  • \( (-\infty, -3) \)
  • \( (-1, 2) \)
It's important to understand that these solutions exclude the interval endpoints, which arose from the critical points. Hence, \( x = -3, x = -1, \) and \( x = 2 \) are not part of the solution, because these are points where the expression either becomes undefined or equals zero.

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