Sove `((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 the denominator

The numerator can be factored as follows:

1. Recognize that \(x^2 - 2x + 1\) is a perfect square:
\[ x^2 - 2x + 1 = (x - 1)^2. \]

Thus, the numerator becomes:
\[ (x + 2)(x - 1)^2. \]

For the denominator, rewrite it:
\[ -4 + 3x - x^2 = - (x^2 - 3x + 4). \]
Now, we will analyze the quadratic \(x^2 - 3x + 4\).

### Step 2: Analyze the denominator

To determine the sign of \(x^2 - 3x + 4\), we can find its discriminant:
\[ D = b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot 4 = 9 - 16 = -7. \]
Since the discriminant is negative, the quadratic \(x^2 - 3x + 4\) has no real roots and is always positive (as the coefficient of \(x^2\) is positive).

Thus,
\[ -(x^2 - 3x + 4) < 0 \text{ for all } x. \]

### Step 3: Rewrite the inequality

Now we can rewrite the inequality:
\[ \frac{(x + 2)(x - 1)^2}{-(x^2 - 3x + 4)} \geq 0 \implies (x + 2)(x - 1)^2 \leq 0. \]

### Step 4: Find the critical points

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

### Step 5: Test intervals around the critical points

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

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

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

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

### Step 6: Determine the solution set

The expression \((x + 2)(x - 1)^2\) is less than or equal to zero in the interval \((- \infty, -2]\) and at \(x = 1\) (where it is zero). Therefore, the solution set is:
\[ x \in (-\infty, -2] \cup \{1\}. \]

### Final Answer

Thus, the solution to the inequality is:
\[ x \in (-\infty, -2] \cup \{1\}. \]