Solve the inequality:
$$
\frac{(x + 2)(x^2 - 2x + 1)}{4 + 3x - x^2} \geq 0
$$
and determine the correct range for $$ x $$ from the given options.

Question

Solve the inequality:  
$$
\frac{(x + 2)(x^2 - 2x + 1)}{4 + 3x - x^2} \geq 0
$$  
and determine the correct range for $$ x $$ from the given options.

Accepted Answer

To solve the inequality  
$$
\frac{(x + 2)(x^2 - 2x + 1)}{4 + 3x - x^2} \geq 0,
$$  
we start by analyzing the numerator and denominator separately.

The numerator is given as:  
$$
(x + 2)(x^2 - 2x + 1).
$$  
Factoring $$ x^2 - 2x + 1 $$, we get:  
$$
(x + 2)(x - 1)^2.
$$

The denominator is given as:  
$$
4 + 3x - x^2.
$$  
Rewriting it, we have:  
$$
-(x^2 - 3x - 4).
$$  
Factoring $$ x^2 - 3x - 4 $$, we get:  
$$
-(x - 4)(x + 1).
$$

Thus, the inequality becomes:  
$$
\frac{(x + 2)(x - 1)^2}{-(x - 4)(x + 1)} \geq 0.
$$

Next, identify the critical points where the expression equals zero or is undefined. These are:  
$$ x = -2, x = -1, x = 1, x = 4 $$.

Now analyze the sign of the expression in each interval formed by these points:  
- For $$ (-\infty, -2) $$, the expression is positive.  
- For $$ (-2, -1) $$, the expression is negative.  
- For $$ (-1, 1) $$, the expression is positive.  
- For $$ (1, 4) $$, the expression is positive.  
- For $$ (4, \infty) $$, the expression is negative.

Considering the inequality $$ \geq 0 $$, we include intervals where the expression is non-negative:  
$$ x < -2 $$ or $$ -1 < x < 4 $$.

Thus, the solution to the inequality is:  
$$ x < -2 $$ or $$ -1 < x < 4 $$.

Answer

Answer

Answer