To determine the solution set for the inequality [tex]\(5(x-2)(x+4) > 0\)[/tex], follow these steps:
1. Identify the roots of the quadratic expression:
Start by setting the expression inside the inequality to zero:
[tex]\[
5(x-2)(x+4) = 0
\][/tex]
Solving this equation, we get two roots:
[tex]\[
x = 2 \quad \text{and} \quad x = -4
\][/tex]
2. Divide the number line into regions based on the roots:
The roots [tex]\(x = 2\)[/tex] and [tex]\(x = -4\)[/tex] divide the number line into three intervals:
[tex]\[
\text{Region 1: } x < -4
\][/tex]
[tex]\[
\text{Region 2: } -4 < x < 2
\][/tex]
[tex]\[
\text{Region 3: } x > 2
\][/tex]
3. Determine the sign of the expression [tex]\(5(x-2)(x+4)\)[/tex] in each region:
- For Region 1 [tex]\( (x < -4)\)[/tex]:
- Pick a test point [tex]\( x = -5 \)[/tex]
- Substitute [tex]\( x = -5 \)[/tex] into the expression:
[tex]\[
5(-5-2)(-5+4) = 5(-7)(-1) = 35 \quad (\text{Positive})
\][/tex]
- For Region 2 [tex]\( (-4 < x < 2)\)[/tex]:
- Pick a test point [tex]\( x = 0 \)[/tex]
- Substitute [tex]\( x = 0 \)[/tex] into the expression:
[tex]\[
5(0-2)(0+4) = 5(-2)(4) = -40 \quad (\text{Negative})
\][/tex]
- For Region 3 [tex]\( (x > 2)\)[/tex]:
- Pick a test point [tex]\( x = 3 \)[/tex]
- Substitute [tex]\( x = 3 \)[/tex] into the expression:
[tex]\[
5(3-2)(3+4) = 5(1)(7) = 35 \quad (\text{Positive})
\][/tex]
4. Combine the regions where the inequality is satisfied:
We are looking for where [tex]\(5(x-2)(x+4) > 0\)[/tex]. From the above analysis:
- The expression is positive in Region 1 [tex]\(x < -4\)[/tex].
- The expression is negative in Region 2 [tex]\(-4 < x < 2\)[/tex].
- The expression is positive in Region 3 [tex]\(x > 2\)[/tex].
Therefore, the solution set where the inequality [tex]\(5(x-2)(x+4) > 0\)[/tex] holds is:
[tex]\[
\{x \mid x < -4 \text{ or } x > 2\}
\][/tex]
Thus, the correct answer is:
[tex]\[
\boxed{\{x \mid x < -4 \text{ or } x > 2\}}
\][/tex]
