Solve `((x+2)(x^2-2x+1))/(-4 +3x-x^2) ge 0 `

Explanation

To solve the inequality \(\frac{(x+2)(x^2-2x+1)}{-4 + 3x - x^2} \geq 0\), we will follow these steps:

### Step 1: Factor the numerator and denominator

1. **Numerator**:
\[ (x+2)(x^2 - 2x + 1) = (x+2)(x-1)^2 \]
Here, \(x^2 - 2x + 1\) can be factored as \((x-1)(x-1)\) or \((x-1)^2\).

2. **Denominator**:
\[ -4 + 3x - x^2 = -(x^2 - 3x + 4) \]
We can rewrite it as:
\[ -(x^2 - 3x + 4) \]
To factor \(x^2 - 3x + 4\), we can check the discriminant:
\[ D = b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot 4 = 9 - 16 = -7 \]
Since the discriminant is negative, \(x^2 - 3x + 4\) has no real roots and is always positive. Therefore, \(- (x^2 - 3x + 4)\) is always negative.

### Step 2: Rewrite the inequality

Now we can rewrite the inequality:
\[ \frac{(x+2)(x-1)^2}{-(x^2 - 3x + 4)} \geq 0 \]
This simplifies to:
\[ -(x+2)(x-1)^2 \geq 0 \]
or equivalently:
\[ (x+2)(x-1)^2 \leq 0 \]

### Step 3: Identify critical points

The critical points occur when the numerator is zero:
1. \(x + 2 = 0 \Rightarrow x = -2\)
2. \((x - 1)^2 = 0 \Rightarrow x = 1\)

### Step 4: Test intervals

We will test the sign of \((x + 2)(x - 1)^2\) in the intervals determined by the critical points:
- \( (-\infty, -2) \)
- \( (-2, 1) \)
- \( (1, \infty) \)

1. **Interval \((- \infty, -2)\)**: Choose \(x = -3\)
\[ (-3 + 2)(-3 - 1)^2 = (-1)(16) < 0 \]
(Negative)

2. **Interval \((-2, 1)\)**: Choose \(x = 0\)
\[ (0 + 2)(0 - 1)^2 = (2)(1) > 0 \]
(Positive)

3. **Interval \((1, \infty)\)**: Choose \(x = 2\)
\[ (2 + 2)(2 - 1)^2 = (4)(1) > 0 \]
(Positive)

### Step 5: Determine the solution set

From our tests, we find:
- The expression is negative in the interval \((- \infty, -2)\).
- The expression is zero at \(x = -2\) and \(x = 1\) (since \((x-1)^2 = 0\)).
- The expression is positive in the intervals \((-2, 1)\) and \((1, \infty)\).

Thus, the solution to the inequality \((x + 2)(x - 1)^2 \leq 0\) is:
\[ x \in (-\infty, -2] \cup \{1\} \]

### Final Answer
\[ \boxed{(-\infty, -2] \cup \{1\}} \]