To solve the inequality:

\[
(2x - 3)(x - 3)(x - 1)^2 \geq 0
\]

### Step 1: Find the roots
To find the roots, set each factor equal to zero:

- \( 2x - 3 = 0 \) ⟹ \( x = \frac{3}{2} \)
- \( x - 3 = 0 \) ⟹ \( x = 3 \)
- \( (x - 1)^2 = 0 \) ⟹ \( x = 1 \) (with multiplicity 2)

So, the roots are \( x = \frac{3}{2}, x = 3, \) and \( x = 1 \).

### Step 2: Analyze the sign of the expression in intervals
We will test the sign of the expression in the following intervals determined by the roots:

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

To check the sign, let's choose a test point from each interval:

#### Interval 1: \( (-\infty, 1) \)
Choose \( x = 0 \):
\[
(2(0) - 3)(0 - 3)(0 - 1)^2 = (-3)(-3)(1) = 9 > 0
\]
The expression is positive in this interval.

#### Interval 2: \( (1, \frac{3}{2}) \)
Choose \( x = 1.2 \):
\[
(2(1.2) - 3)(1.2 - 3)(1.2 - 1)^2 = (-0.6)(-1.8)(0.04) = 0.0432 > 0
\]
The expression is positive in this interval.

#### Interval 3: \( \left( \frac{3}{2}, 3 \right) \)
Choose \( x = 2 \):
\[
(2(2) - 3)(2 - 3)(2 - 1)^2 = (1)(-1)(1) = -1 < 0
\]
The expression is negative in this interval.

#### Interval 4: \( (3, \infty) \)
Choose \( x = 4 \):
\[
(2(4) - 3)(4 - 3)(4 - 1)^2 = (5)(1)(9) = 45 > 0
\]
The expression is positive in this interval.

### Step 3: Consider the points where the expression is zero
The expression equals zero at \( x = \frac{3}{2}, x = 3, \) and \( x = 1 \). Since the inequality is \( \geq 0 \), these points are included in the solution.

### Step 4: Combine the intervals
From the analysis, the expression is positive or zero in the following intervals:

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

### Final Solution:
\[
(-\infty, 1] \cup \left( 1, \frac{3}{2} \right] \cup [3, \infty)
\]

Would you like more details or have any questions?

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Here are some related questions to explore further:
1. What are the key differences between solving quadratic and cubic inequalities?
2. How does the multiplicity of roots affect the sign of a function?
3. Why does squaring a factor like \( (x - 1)^2 \) not change the sign in any interval?
4. How do inequalities differ when there is a strict inequality (e.g., \( > \) instead of \( \geq \))?
5. Can the sign analysis technique be extended to inequalities involving higher degree polynomials?

**Tip:** Always consider the multiplicity of roots—if the multiplicity is even, the factor does not change sign at the root!

Learn how to solve the polynomial inequality (2x-3)(x-3)(x-1)^2 ≥ 0 with a detailed step-by-step solution. This problem covers key algebraic concepts, such as finding roots, analyzing intervals, and understanding the impact of root multiplicity. Suitable for students in Grades 9-12.

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