Thank you for the clarification! If the expression is meant to be part of an inequality, then we should consider the inequality and solve for the values of $x$ that satisfy it.

Let's assume the inequality involves the expression provided:

$\frac{(2x - 4)(x + 3)}{x - 2} > 0$

We want to find the values of $x$ for which the expression is greater than zero. Here’s how we can solve this inequality step-by-step:

Step-by-Step Solution

1. Find the critical points: Determine the values of $x$ where the expression equals zero or where it is undefined. These points divide the number line into intervals where the sign of the expression might change.

   • Numerator is zero: $(2x - 4)(x + 3) = 0$

     $2x - 4 = 0 \quad \text{or} \quad x + 3 = 0$ $x = 2 \quad \text{or} \quad x = -3$

   • Denominator is zero: $x - 2 = 0$

     $x = 2$

   So, the critical points are $x = -3$, $x = 2$, and the point where the expression is undefined is also $x = 2$.

2. Determine the sign of the expression in each interval: The critical points $-3$ and $2$ divide the number line into three intervals: $(- \infty, -3)$, $(-3, 2)$, and $(2, \infty)$. We need to test a point in each interval to determine whether the expression is positive or negative in that interval.

   • Interval $(- \infty, -3)$: Choose $x = -4$

     $\frac{(2(-4) - 4)((-4) + 3)}{-4 - 2} = \frac{(-8 - 4)(-1)}{-6} = \frac{(-12)(-1)}{-6} = \frac{12}{-6} = -2$

     The expression is negative in this interval.

   • Interval $(-3, 2)$: Choose $x = 0$

     $\frac{(2(0) - 4)(0 + 3)}{0 - 2} = \frac{(-4)(3)}{-2} = \frac{-12}{-2} = 6$

     The expression is positive in this interval.

   • Interval $(2, \infty)$: Choose $x = 3$

     $\frac{(2(3) - 4)(3 + 3)}{3 - 2} = \frac{(6 - 4)(6)}{1} = \frac{(2)(6)}{1} = 12$

     The expression is positive in this interval.

3. Write the solution based on the signs:

   The inequality $\frac{(2x - 4)(x + 3)}{x - 2} > 0$ is positive on the intervals where the product is positive. From our testing:

   • It is positive in the intervals $(-3, 2)$ and $(2, \infty)$.

   • We must exclude $x = 2$ because the expression is undefined at this point.

4. Solution in interval notation:

   The solution set for the inequality $\frac{(2x - 4)(x + 3)}{x - 2} > 0$ is:

   $(-3, 2) \cup (2, \infty)$

Conclusion

The inequality $\frac{(2x - 4)(x + 3)}{x - 2} > 0$ is satisfied for $x$ in the intervals $(-3, 2)$ and $(2, \infty)$.

Would you like more details on any specific step, or do you have any questions?

Relative Questions for Further Exploration:

1. How would the solution change if the inequality were $\frac{(2x - 4)(x + 3)}{x - 2} < 0$?
2. How do you handle inequalities that include equalities, like $\geq$ or $\leq$?
3. What happens to the solution if the numerator or denominator has different signs?
4. Can you graph the inequality to visually confirm the intervals?
5. How do you solve inequalities involving absolute values?

Tip:

When solving inequalities, it's helpful to create a number line and test points in each interval. This approach ensures you accurately determine where the expression is positive or negative.