We are tasked with solving the inequality:

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

### Step 1: Find the critical points
The inequality will change sign at the points where any factor becomes zero. Let's set each factor equal to zero:

\[
(х - 1)^2 = 0 \implies х = 1
\]
\[
(х + 2) = 0 \implies х = -2
\]
\[
(х - 3)^3 = 0 \implies х = 3
\]

Thus, the critical points are \( х = 1 \), \( х = -2 \), and \( х = 3 \).

### Step 2: Determine the sign of each factor in each interval
We will divide the real line into intervals based on these critical points: \( (-\infty, -2) \), \( (-2, 1) \), \( (1, 3) \), and \( (3, \infty) \).

Now, let's check the sign of the product in each of these intervals:

1. **For \( х \in (-\infty, -2) \):**
   \[
   (х - 1)^2 > 0, \quad (х + 2) < 0, \quad (х - 3)^3 < 0
   \]
   The product \( (х - 1)^2(х + 2)(х - 3)^3 \) is \( (+)(-)(-) = + \), so it's positive in this interval.

2. **For \( х \in (-2, 1) \):**
   \[
   (х - 1)^2 > 0, \quad (х + 2) > 0, \quad (х - 3)^3 < 0
   \]
   The product \( (х - 1)^2(х + 2)(х - 3)^3 \) is \( (+)(+)(-) = - \), so it's negative in this interval.

3. **For \( х \in (1, 3) \):**
   \[
   (х - 1)^2 > 0, \quad (х + 2) > 0, \quad (х - 3)^3 < 0
   \]
   Again, the product \( (х - 1)^2(х + 2)(х - 3)^3 \) is \( (+)(+)(-) = - \), so it's negative in this interval.

4. **For \( х \in (3, \infty) \):**
   \[
   (х - 1)^2 > 0, \quad (х + 2) > 0, \quad (х - 3)^3 > 0
   \]
   The product \( (х - 1)^2(х + 2)(х - 3)^3 \) is \( (+)(+)(+) = + \), so it's positive in this interval.

### Step 3: Check behavior at the critical points
- At \( х = -2 \), the factor \( (х + 2) = 0 \), so the product is zero.
- At \( х = 1 \), the factor \( (х - 1)^2 = 0 \), so the product is zero.
- At \( х = 3 \), the factor \( (х - 3)^3 = 0 \), so the product is zero.

### Step 4: Conclusion
The inequality holds when the product is greater than or equal to zero. From our analysis, this occurs in the intervals \( (-\infty, -2] \cup [1, 3] \cup [3, \infty) \).

Thus, the solution is:
\[
х \in (-\infty, -2] \cup [1, \infty)
\]

Would you like more details or have any questions?

Here are five related questions you can explore:
1. What happens if we modify the inequality to strict \( > \)?
2. How would the solution change if one of the factors had an odd power?
3. Can you apply a similar method to solve other polynomial inequalities?
4. What is the geometric significance of each critical point in the graph of the function?
5. How does the behavior of even and odd powers affect the sign of the product?

**Tip:** Always check behavior at the critical points separately when solving polynomial inequalities.

Explore the step-by-step solution to the polynomial inequality (x-1)^2(x+2)(x-3)^3 ≥ 0. This problem demonstrates the use of sign analysis and critical points to solve inequalities, suitable for students in Grades 10-12.

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