1. Set to Zero and Factor: Ensure the inequality is in the form $R(x) \geq 0$, $R(x) > 0$, etc., where $R(x)$ is a single rational expression.
2. Find the Critical Points (Roots):
Set the numerator equal to zero: $(x-1)^2(x-2) = 0$. This gives $x=1$ and $x=2$ (the roots).
Set the denominator equal to zero: $(x-3)^3 = 0$. This gives $x=3$.
Note: At $x=1$ and $x=2$, the expression is zero. At $x=3$, the expression is undefined (U.D.).
3. Plot on the Number Line: Write the $x$ values in increasing order: $-\infty \dots 1 \dots 2 \dots 3 \dots \infty$.
4. Determine the Signs of the Intervals:
Start Right: Check the sign of the expression for a value greater than the largest root (e.g., $x=4$). The sign after the last root (3) is typically positive ($+$).
Change Signs based on Power:
If a root occurs an odd number of times (like $x=3$ from $(x-3)^3$, and $x=2$ from $(x-2)^1$), the sign changes across that root.
If a root occurs an even number of times (like $x=1$ from $(x-1)^2$), the sign stays the same across that root.
| Interval | Test Point | Exponent Status | Sign Change? |
| $(3, \infty)$ | $x=4$ | (Start) | $+$ |
| $(2, 3)$ | $x=2.5$ | $x=3$ is odd exponent | Sign changes to $-$ |
| $(1, 2)$ | $x=1.5$ | $x=2$ is odd exponent | Sign changes to $+$ |
| $(-\infty, 1)$ | $x=0$ | $x=1$ is even exponent | Sign stays $+$ |
5. Write the Solution Set: Use your sign analysis to answer the specific inequality request:
| Inequality | Solution Set | Explanation |
| $\frac{(x-1)^2(x-2)}{(x-3)^3} < 0$ | $x \in (2, 3)$ | Where the expression is negative ($-$). |
| $\frac{(x-1)^2(x-2)}{(x-3)^3} \leq 0$ | $x \in [2, 3) \cup \{1\}$ | Negative ($-$) or equal to zero. Zero at $x=1$ and $x=2$. $x=3$ is always excluded (U.D.). |
| $\frac{(x-1)^2(x-2)}{(x-3)^3} > 0$ | $x \in (-\infty, 1) \cup (1, 2) \cup (3, \infty)$ | Where the expression is positive ($+$). |
| $\frac{(x-1)^2(x-2)}{(x-3)^3} \geq 0$ | $x \in (-\infty, 2] \cup (3, \infty)$ | Positive ($+$) or equal to zero. Note: The set $(-\infty, 1) \cup \{1\} \cup (1, 2]$ simplifies to $(-\infty, 2]$. $x=3$ is excluded. |