$$ \rightarrow $$ Reselue into partial fraction $$ \frac{2-4 x}{x\left(x^{2}-x-2\right)} $$

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To decompose the expression $$ \frac{2-4x}{x(x^2 - x - 2)} $$ into partial fractions, we first need to factor the denominator completely.

The quadratic $$ x^2 - x - 2 $$ can be factored as follows:

$$
x^2 - x - 2 = (x - 2)(x + 1)
$$

Thus, the complete factorization of the denominator is:

$$
x(x - 2)(x + 1)
$$

Now, we can express the fraction as a sum of partial fractions:

$$
\frac{2 - 4x}{x(x - 2)(x + 1)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 1}
$$

where $$ A $$, $$ B $$, and $$ C $$ are constants that we need to determine.

Next, we multiply both sides by the denominator $$ x(x - 2)(x + 1) $$ to eliminate the fraction:

$$
2 - 4x = A(x - 2)(x + 1) + B(x)(x + 1) + C(x)(x - 2)
$$

Now, we will expand the right-hand side:

1. Expanding $$ A(x - 2)(x + 1) $$:
   $$
   A(x^2 - x - 2)
   $$

2. Expanding $$ B(x)(x + 1) $$:
   $$
   B(x^2 + x)
   $$

3. Expanding $$ C(x)(x - 2) $$:
   $$
   C(x^2 - 2x)
   $$

Combining these, we have:

$$
2 - 4x = A(x^2 - x - 2) + B(x^2 + x) + C(x^2 - 2x)
$$

Now, we can combine like terms:

$$
2 - 4x = (A + B + C)x^2 + (-A + B - 2C)x - 2A
$$

Next, we equate the coefficients from both sides of the equation:

1. For $$ x^2 $$: $$ A + B + C = 0 $$
2. For $$ x $$: $$ -A + B - 2C = -4 $$
3. For the constant term: $$ -2A = 2 $$

From the third equation, we can solve for $$ A $$:

$$
-2A = 2 \implies A = -1
$$

Substituting $$ A = -1 $$ into the first equation:

$$
-1 + B + C = 0 \implies B + C = 1 \quad \text{(1)}
$$

Now substituting $$ A = -1 $$ into the second equation:

$$
-(-1) + B - 2C = -4 \implies 1 + B - 2C = -4 \implies B - 2C = -5 \quad \text{(2)}
$$

Now we have a system of equations:

1. $$ B + C = 1 $$ (1)
2. $$ B - 2C = -5 $$ (2)

We can solve this system. From equation (1), we can express $$ B $$ in terms of $$ C $$:

$$
B = 1 - C
$$

Substituting this into equation (2):

$$
(1 - C) - 2C = -5
$$
$$
1 - 3C = -5
$$
$$
-3C = -6 \implies C = 2
$$

Now substituting $$ C = 2 $$ back into equation (1):

$$
B + 2 = 1 \implies B = -1
$$

Now we have:

$$
A = -1, \quad B = -1, \quad C = 2
$$

Thus, the partial fraction decomposition is:

$$
\frac{2 - 4x}{x(x - 2)(x + 1)} = \frac{-1}{x} + \frac{-1}{x - 2} + \frac{2}{x + 1}
$$

So, the final result is:

$$
\frac{2 - 4x}{x(x - 2)(x + 1)} = -\frac{1}{x} - \frac{1}{x - 2} + \frac{2}{x + 1}
$$
The partial fraction decomposition of $$ \frac{2-4x}{x(x^2 - x - 2)} $$ is $$ -\frac{1}{x} - \frac{1}{x - 2} + \frac{2}{x + 1} $$.

\( \rightarrow \) Reselue into partial fraction \( \frac{2-4 x}{x\left(x^{2}-x-2\right)} \)

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