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]
