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!