Use a sign chart to solve [tex](2x + 3)(x + 8) \geq 0[/tex].

A. [tex](-\infty, -8) \cup \left( -\frac{3}{2}, \infty \right)[/tex]
B. [tex](-\infty, -8] \cup \left[-\frac{3}{2}, \infty \right)[/tex]
C. [tex](-8, -\frac{3}{2})[/tex]
D. [tex][-8, -\frac{3}{2}][/tex]



Answer :

Sure, let's solve the inequality [tex]\((2 x + 3)(x + 8) \geq 0\)[/tex] step by step using a sign chart.

### Step 1: Identify the Critical Points

First, we find the values of [tex]\(x\)[/tex] where the expression is zero, because these values split the number line into intervals that we can test.

[tex]\[ (2x + 3)(x + 8) = 0 \][/tex]

Set each factor equal to zero:

1. [tex]\(2x + 3 = 0 \implies x = -\frac{3}{2}\)[/tex]
2. [tex]\(x + 8 = 0 \implies x = -8\)[/tex]

So, the critical points are [tex]\(x = -\frac{3}{2}\)[/tex] and [tex]\(x = -8\)[/tex].

### Step 2: Create Intervals

The critical points divide the number line into three intervals:
1. [tex]\( (-\infty, -8) \)[/tex]
2. [tex]\( (-8, -\frac{3}{2}) \)[/tex]
3. [tex]\( (-\frac{3}{2}, \infty) \)[/tex]

### Step 3: Test the Sign of Each Interval

We need to check the sign of [tex]\((2x + 3)(x + 8)\)[/tex] within each interval.

#### Interval 1: [tex]\( (-\infty, -8) \)[/tex]

Choose a test point, such as [tex]\(x = -9\)[/tex]:

[tex]\[ (2(-9) + 3)(-9 + 8) = (-18 + 3)(-1) = (-15)(-1) = 15 \quad (\text{positive}) \][/tex]

#### Interval 2: [tex]\((-8, -\frac{3}{2}) \)[/tex]

Choose a test point, such as [tex]\(x = -5\)[/tex]:

[tex]\[ (2(-5) + 3)(-5 + 8) = (-10 + 3)(3) = (-7)(3) = -21 \quad (\text{negative}) \][/tex]

#### Interval 3: [tex]\((- \frac{3}{2}, \infty) \)[/tex]

Choose a test point, such as [tex]\(x = 0\)[/tex]:

[tex]\[ (2(0) + 3)(0 + 8) = (3)(8) = 24 \quad (\text{positive}) \][/tex]

Based on these calculations, we see the sign pattern is:

- Positive in [tex]\( (-\infty, -8) \)[/tex]
- Negative in [tex]\( (-8, -\frac{3}{2}) \)[/tex]
- Positive in [tex]\( (-\frac{3}{2}, \infty) \)[/tex]

### Step 4: Include the Critical Points

Since the inequality is [tex]\(\geq 0\)[/tex], we must include the critical points where the expression is zero ([tex]\(x = -\frac{3}{2}\)[/tex] and [tex]\(x = -8\)[/tex]):

- [tex]\((2x + 3)(x + 8) = 0\)[/tex] at [tex]\(x = -\frac{3}{2}\)[/tex] and [tex]\(x = -8\)[/tex]

### Step 5: Combine the Intervals

The solution set for the inequality includes the intervals where the expression is non-negative, including the critical points:

[tex]\[ (-\infty, -8] \cup \left[-\frac{3}{2}, \infty\right) \][/tex]

So, the solution to the inequality [tex]\((2 x + 3)(x + 8) \geq 0\)[/tex] is:

[tex]\[ (-\infty, -8] \cup \left[-\frac{3}{2}, \infty\right) \][/tex]

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