Problem 35 - Beginning and Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The solution set is \ \[-\frac{3}{2}, \frac{1}{3}] \cup [4, \infty)\].

Step 1 - Find the zeros of the inequality

To solve the inequality (x-4)(2x+3)(3x-1) \geq 0, we need to find the values of \( x \) that make each factor equal to zero. These are the points where the inequality could change sign. Solve for each factor: 1) \( x-4 = 0 \implies x = 4 \) 2) \( 2x+3 = 0 \implies x = -\frac{3}{2} \) 3) \( 3x-1 = 0 \implies x = \frac{1}{3} \)

Step 2 - Determine the intervals

The zeros divide the real number line into four intervals: 1) \(( -\infty, -\frac{3}{2} )\) 2) \((-\frac{3}{2}, \frac{1}{3})\) 3) \( (\frac{1}{3}, 4 )\) 4) \((4, \infty)\)

Step 3 - Test the sign in each interval

Choose a test point from each interval to determine if the product (x-4)(2x+3)(3x-1) \geq 0 is positive or negative in that interval: 1) For \( x = -2 \) in \(( -\infty, -\frac{3}{2} ) \), (\(x-4 \))\(=-2-4=-6 \) (negative), (\(2x+3 \)) \(=2(-2)+3=-4+3=-1 \) (negative), (\(3x-1 \)) \(=3(-2)-1=-6-1=-7 \) (negative), thus the product is \(- * - * - = -\) (negative). 2) For \( x = 0 \) in \((-\frac{3}{2}, \frac{1}{3} )\), (\(x-4 \))\(=0-4=-4\) (negative), (\(2x+3 \))\(=2(0)+3=3\) (positive), (\(3x-1\))\(=3(0)-1=-1\) (negative), thus the product is \(- * + * - = +\) (positive). 3) For \( x = 1 \) in \(( \frac{1}{3}, 4 )\), (\(x-4 \))\(=1-4=-3\) (negative), (\(2x+3 \)) \(=2(1)+3=5\) (positive), (\(3x-1 \)) \(= 3(1)-1=2\) (positive), thus the product is \(- * + * + = - \) (negative). 4) For \( x = 5 \) in \((4, \infty)\), (\(x-4 \))\(=5-4=1\) (positive), (\(2x+3 \))\(=2(5)+3=10+3=13\) (positive), (\(3x-1 \))\(=3(5)-1=15-1=14\) (positive), thus the product is \(+ * + * + = +\) (positive).

Step 4 - Determine the intervals that satisfy the inequality

Identify the intervals where the product is greater than or equal to zero: From the test points, the inequality is satisfied in the intervals: \ \ \( -\frac{3}{2}, \frac{1}{3} ) \cup (4, \infty)\).

Step 5 - Include the zeros

Since the inequality is \( \geq 0\), include the zeros where the product is equal to zero: \[-\frac{3}{2}, \frac{1}{3} ] \cup [4, \infty)\].

Step 6 - Graph the solution

Graph the intervals on a number line. Include brackets at \(-\frac{3}{2}\), and \(\frac{1}{3}\) and \(4\).

Key Concepts

Zeros of Polynomial

When solving polynomial inequalities, the first step is to find the zeros of the polynomial. These are the values of \( x \) that make each factor of the polynomial equal to zero. For the inequality \((x-4)(2x+3)(3x-1) \geq 0 \), we solve the following equations to find the zeros:
\( x - 4 = 0 \rightarrow x = 4 \)
\( 2x + 3 = 0 \rightarrow x = -\frac{3}{2} \)
\( 3x - 1 = 0 \rightarrow x = \frac{1}{3} \).
These zeros divide the number line into intervals where the sign of the polynomial can be tested. Identifying these zeros is crucial as the inequality could change its sign at these points.

Inequality Intervals

Once you have the zeros, the next step is to determine the intervals on the number line created by these zeros. For our polynomial \((x-4)(2x+3)(3x-1) \geq 0 \), the zeros \( x = 4 \), \( x = -\frac{3}{2} \), and \( x = \frac{1}{3} \) split the real number line into four distinct intervals:

• \( ( -\infty, -\frac{3}{2} ) \)


• \( ( -\frac{3}{2},\frac{1}{3} ) \)


• \( ( \frac{1}{3}, 4 ) \)


• \( ( 4, \infty ) \)

These intervals are where we’ll test the sign of the polynomial. These test points will reveal if the product of the factors is positive or negative in the interval.

Test Points in Intervals

For each interval determined by the zeros, select a test point and plug it into the polynomial to check the sign of the product. Here's how it works for our example \((x-4)(2x+3)(3x-1) \geq 0 \):

• For \(x = -2\) in \( ( -\infty, -\frac{3}{2} ) \): The product \( (-6)*(-1)*(-7) = - \) (negative).


• For \(x = 0\) in \( ( -\frac{3}{2},\frac{1}{3} ) \): The product \( (-4)*(3)*(-1) = + \) (positive).


• For \(x = 1\) in \( ( \frac{1}{3}, 4 ) \): The product \( (-3)*(5)*(2) = - \) (negative).


• For \(x = 5\) in \( ( 4, \infty ) \): The product \( (1)*(13)*(14) = + \) (positive).

By testing these points, we can identify which intervals satisfy the original inequality.

Graphing Solution Sets

After determining the intervals where the inequality holds true, graph the solution on a number line. From our test points, the inequality \((x-4)(2x+3)(3x-1) \geq 0 \) is satisfied for the intervals \( ( -\frac{3}{2},\frac{1}{3} ) \) and \( (4, \infty) \). Since the inequality includes \(\geq \) (greater than or equal to), we also include the zeros: \( -\frac{3}{2}, \frac{1}{3}, \) and \(4\).
Thus, the solution set is \[ -\frac{3}{2}, \frac{1}{3} ] \cup [4, \infty) \]. When graphing, use brackets at \( -\frac{3}{2}, \frac{1}{3}, \) and \(4\) to show these points are included in the solution set. This visual representation helps to clearly see where the inequality holds true on the number line.