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Let's break down the information step by step:

We are given a function:

\[
P(x) = x(x + 2)(x - 1)
\]

### Step 1: Identify Critical Points
First, we find the roots of \(P(x)\), which are the values where the function equals zero:

- \(x = 0\)
- \(x + 2 = 0 \Rightarrow x = -2\)
- \(x - 1 = 0 \Rightarrow x = 1\)

Thus, the critical points where the sign of the function may change are at \(x = -2\), \(x = 0\), and \(x = 1\).

### Step 2: Sign Analysis

We now perform a sign analysis on the intervals determined by the critical points: \((- \infty, -2)\), \((-2, 0)\), \((0, 1)\), and \((1, \infty)\).

1. **Interval \((- \infty, -2)\)**:
   - \(x < -2\), so \(x\) is negative.
   - \(x + 2 < 0\), so \(x + 2\) is negative.
   - \(x - 1 < 0\), so \(x - 1\) is negative.
   - The product \(x(x + 2)(x - 1)\) is \((-)(-)(-) = -\), so \(P(x) < 0\).

2. **Interval \((-2, 0)\)**:
   - \(x > -2\) and \(x < 0\), so \(x\) is negative.
   - \(x + 2 > 0\), so \(x + 2\) is positive.
   - \(x - 1 < 0\), so \(x - 1\) is negative.
   - The product \(x(x + 2)(x - 1)\) is \((-)(+)(-) = +\), so \(P(x) > 0\).

3. **Interval \((0, 1)\)**:
   - \(x > 0\) and \(x < 1\), so \(x\) is positive.
   - \(x + 2 > 0\), so \(x + 2\) is positive.
   - \(x - 1 < 0\), so \(x - 1\) is negative.
   - The product \(x(x + 2)(x - 1)\) is \((+)(+)(-) = -\), so \(P(x) < 0\).

4. **Interval \((1, \infty)\)**:
   - \(x > 1\), so \(x\) is positive.
   - \(x + 2 > 0\), so \(x + 2\) is positive.
   - \(x - 1 > 0\), so \(x - 1\) is positive.
   - The product \(x(x + 2)(x - 1)\) is \((+)(+)(+) = +\), so \(P(x) > 0\).

### Step 3: Analyze at Critical Points
- At \(x = -2\), \(P(x) = 0\) because of the factor \((x + 2)\).
- At \(x = 0\), \(P(x) = 0\) because of the factor \(x\).
- At \(x = 1\), \(P(x) = 0\) because of the factor \((x - 1)\).

### Step 4: Determine Intervals for \(P(x) \geq 0\)
From the sign analysis, \(P(x) \geq 0\) on the following intervals:
\[
[-2, 0] \cup [1, \infty)
\]

Thus, the solution in interval notation is:
\[
\boxed{[-2, 0] \cup [1, \infty)}
\]

---

Would you like further details or have any questions?

Here are 5 related questions to expand on this topic:
1. What is the significance of the critical points in this analysis?
2. How can we graphically represent the sign changes of \(P(x)\)?
3. What is the difference between solving for \(P(x) \geq 0\) versus \(P(x) > 0\)?
4. How does the multiplicity of roots affect the behavior of polynomial functions?
5. How can we apply this method to other polynomial functions?

**Tip:** When analyzing polynomials, the behavior at the roots (whether the function crosses the x-axis or just touches it) depends on the multiplicity of the roots.

Find the intervals where P(x) = x(x + 2)(x - 1) ≥ 0 using sign analysis.

Learn how to solve the polynomial inequality P(x) = x(x + 2)(x - 1) ≥ 0 using sign analysis and critical points. This step-by-step guide covers key algebraic concepts, including the zero product property and sign chart analysis. Suitable for students in Grades 9-11.

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