Problem 34 - Beginning and Intermediate Algebra

Question

Solve each inequality, and graph the solution set. $$ (2 x+1)(3 x-2)(4 x+7)<0 $$

Step-by-Step Solution

Answer

The solution intervals are (-∞, -7/4) and (-1/2, 2/3).

Step 1: Identify the critical points

Set each factor in the inequality (2x+1)(3x-2)(4x+7) < 0 equal to zero and solve for x. Solve for (2x + 1) = 0: 2x + 1 = 0 x = -1/2 Solve for (3x - 2) = 0: 3x - 2 = 0 x = 2/3 Solve for (4x + 7) = 0: 4x + 7 = 0 x = -7/4.

Step 2: Determine intervals to test

The critical points divide the x-axis into four regions: (-∞, -7/4), (-7/4, -1/2), (-1/2, 2/3), and (2/3, ∞).

Step 3: Test each interval

Select a test point from each interval and determine the sign of the product (2x+1)(3x-2)(4x+7): For (-∞, -7/4), select x = -2: (2(-2)+1)(3(-2)-2)(4(-2)+7) = (-3)(-8)(-1) = -24 (negative) For (-7/4, -1/2), select x = -1: (2(-1)+1)(3(-1)-2)(4(-1)+7) = (-1)(-5)(3) = 15 (positive) For (-1/2, 2/3), select x = 0: (2(0)+1)(3(0)-2)(4(0)+7) = (1)(-2)(7) = -14 (negative) For (2/3, ∞), select x = 1: (2(1)+1)(3(1)-2)(4(1)+7) = (3)(1)(11) = 33 (positive)

Step 4: Determine the solution region

The inequality (2x+1)(3x-2)(4x+7) < 0 holds where the product is negative. The solution intervals are: (-∞, -7/4) and (-1/2, 2/3).

Step 5: Graph the solution set

Plot the solution intervals on a number line. Draw open circles at x = -7/4, x = -1/2, and x = 2/3 to indicate that these points are not included in the solution. Shade the regions (-∞, -7/4) and (-1/2, 2/3).

Key Concepts

inequality graphing

To start solving polynomial inequalities, it's helpful to visualize the inequality by graphing. We first need to find the critical points, which are solutions to the equation formed by setting the polynomial expression equal to zero. Once we have the critical points, we can divide the number line into regions. Typically, these points will be marked as open or closed circles, depending on whether the inequality includes the endpoint or not. Shading the correct intervals helps us easily see where the inequality holds true.

critical points

Critical points are values of x where the polynomial equals zero. They are crucial for solving inequalities because they indicate where the product of polynomial factors changes sign. In this example, the polynomial inequality (2x + 1)(3x - 2)(4x + 7) < 0 is given. We find the critical points by setting each factor to zero and solving:

• For (2x + 1) = 0: 2x + 1 = 0, so x = -1/2
• For (3x - 2) = 0: 3x - 2 = 0, so x = 2/3
• For (4x + 7) = 0: 4x + 7 = 0, so x = -7/4

These critical points are then used to establish intervals for testing.

interval testing

Interval testing involves selecting test points within the intervals created by the critical points. This helps determine where the inequality is true. The critical points (-7/4, -1/2, 2/3) divide the number line into four segments:

• (-∞, -7/4)
• (-7/4, -1/2)
• (-1/2, 2/3)
• (2/3, ∞)

For each segment, choose a test point, plug it into the inequality, and check the sign of the product:

• For (-∞, -7/4), test x = -2
• For (-7/4, -1/2), test x = -1
• For (-1/2, 2/3), test x = 0
• For (2/3, ∞), test x = 1

Note the signs of the results for each interval.

solution intervals

Solution intervals are regions where the inequality holds true. After testing each interval, we identify which regions produce a negative product when substituted back into the polynomial. In our case:
For x = -2 in (-∞, -7/4), the product is negative.For x = -1 in (-7/4, -1/2), the product is positive.For x = 0 in (-1/2, 2/3), the product is negative.For x = 1 in (2/3, ∞), the product is positive.

Therefore, the polynomial inequality (2x+1)(3x-2)(4x+7) < 0 holds true in the intervals (-∞, -7/4) and (-1/2, 2/3). Be sure to use open circles around the critical points and shade these intervals on the number line to complete the graph.