Solve the nonlinear inequality. Express the solution using interval notation.

(2*x* − 1)(*x* − 3)2

*All of that over, x* − 5

 < 0

The given inequality is:

\[
\frac{(2x - 1)(x - 3)^2}{x - 5} < 0
\]

### Step 1: Analyze the critical points
To solve the inequality, we first determine the critical points where the expression equals zero or is undefined. These critical points come from the factors in the numerator and denominator.

1. **Numerator**: \( (2x - 1)(x - 3)^2 \)
   - Set \( 2x - 1 = 0 \), which gives \( x = \frac{1}{2} \).
   - Set \( (x - 3)^2 = 0 \), which gives \( x = 3 \).

2. **Denominator**: \( x - 5 \)
   - Set \( x - 5 = 0 \), which gives \( x = 5 \).

### Step 2: Determine intervals
The critical points divide the real number line into several intervals:

1. \( (-\infty, \frac{1}{2}) \)
2. \( \left( \frac{1}{2}, 3 \right) \)
3. \( (3, 5) \)
4. \( (5, \infty) \)

### Step 3: Analyze the sign of the expression on each interval
We will test the sign of the expression in each interval by picking test points.

- For \( (-\infty, \frac{1}{2}) \), pick \( x = 0 \): 
  \[
  \frac{(2(0) - 1)(0 - 3)^2}{0 - 5} = \frac{(-1)(9)}{-5} = \frac{-9}{-5} = \text{positive}.
  \]

- For \( \left( \frac{1}{2}, 3 \right) \), pick \( x = 1 \):
  \[
  \frac{(2(1) - 1)(1 - 3)^2}{1 - 5} = \frac{(1)(4)}{-4} = \frac{4}{-4} = \text{negative}.
  \]

- For \( (3, 5) \), pick \( x = 4 \):
  \[
  \frac{(2(4) - 1)(4 - 3)^2}{4 - 5} = \frac{(7)(1)}{-1} = \frac{7}{-1} = \text{negative}.
  \]

- For \( (5, \infty) \), pick \( x = 6 \):
  \[
  \frac{(2(6) - 1)(6 - 3)^2}{6 - 5} = \frac{(11)(9)}{1} = \text{positive}.
  \]

### Step 4: Combine intervals
We are looking for when the expression is **negative**. Based on the sign analysis, the expression is negative in:

- \( \left( \frac{1}{2}, 3 \right) \)
- \( (3, 5) \)

### Step 5: Check behavior at critical points
- At \( x = \frac{1}{2} \), the numerator is zero, so the entire expression equals zero, but we need it to be strictly less than zero, so \( x = \frac{1}{2} \) is **not included**.
- At \( x = 3 \), \( (x - 3)^2 = 0 \), but because of the square, the factor doesn't change sign, so it's **not included** in the solution.
- At \( x = 5 \), the denominator is zero, making the expression undefined, so \( x = 5 \) is **not included**.

### Final Solution:
The solution in interval notation is:

\[
\left( \frac{1}{2}, 3 \right) \cup (3, 5)
\]

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Would you like more details or clarifications on any step? Here are a few related questions to deepen your understanding:

1. How do you solve rational inequalities in general?
2. Why isn't \( x = 3 \) included in the solution, even though \( (x - 3)^2 = 0 \)?
3. What is the significance of analyzing the denominator in rational inequalities?
4. How do you determine the sign of an expression over an interval?
5. What happens when the critical point makes the numerator zero?

**Tip:** Always remember to check critical points and verify whether they are included in the solution set based on the inequality sign.

Solve the nonlinear inequality. Express the solution using interval notation. (2x - 1)(x - 3)^2 / (x - 5) < 0

Learn how to solve the nonlinear inequality (2x - 1)(x - 3)^2 / (x - 5) < 0. This guide covers key mathematical concepts such as finding critical points, performing sign analysis, and expressing the solution in interval notation. Suitable for high school students in Grades 10-12.

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