Resolve into partial fractions :
`(6x^(2)+5x-2)/(2x^(3)-x^(2)-x)`.

Answer

To resolve the expression \(\frac{6x^2 + 5x - 2}{2x^3 - x^2 - x}\) into partial fractions, we will follow these steps:

### Step 1: Factor the Denominator
First, we need to factor the denominator \(2x^3 - x^2 - x\).

\[ 2x^3 - x^2 - x = x(2x^2 - x - 1) \]

Next, we will factor \(2x^2 - x - 1\). We can do this by finding two numbers that multiply to \(-2\) (the product of \(2\) and \(-1\)) and add to \(-1\) (the coefficient of \(x\)). The numbers \(-2\) and \(1\) work.

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

Thus, the complete factorization of the denominator is:

\[ 2x^3 - x^2 - x = x(2x + 1)(x - 1) \]

### Step 2: Set Up the Partial Fraction Decomposition
Now we can express the fraction as a sum of partial fractions:

\[ \frac{6x^2 + 5x - 2}{x(2x + 1)(x - 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{2x + 1} \]

### Step 3: Clear the Denominator
Multiply both sides by the denominator \(x(2x + 1)(x - 1)\):

\[ 6x^2 + 5x - 2 = A(2x + 1)(x - 1) + B(x)(2x + 1) + C(x)(x - 1) \]

### Step 4: Expand the Right Side
Now we will expand the right side:

1. For \(A(2x + 1)(x - 1)\):
\[ A(2x^2 - 2x + x - 1) = A(2x^2 - x - 1) \]

2. For \(B(x)(2x + 1)\):
\[ B(2x^2 + x) \]

3. For \(C(x)(x - 1)\):
\[ C(x^2 - x) \]

Combining these, we have:

\[ 6x^2 + 5x - 2 = (2A + 2B + C)x^2 + (-A + B - C)x - A \]

### Step 5: Compare Coefficients
Now we can compare coefficients from both sides:

1. Coefficient of \(x^2\):
\[ 2A + 2B + C = 6 \quad \text{(1)} \]

2. Coefficient of \(x\):
\[ -A + B - C = 5 \quad \text{(2)} \]

3. Constant term:
\[ -A = -2 \quad \Rightarrow \quad A = 2 \quad \text{(3)} \]

### Step 6: Solve the System of Equations
Substituting \(A = 2\) into equations (1) and (2):

From (1):
\[ 2(2) + 2B + C = 6 \quad \Rightarrow \quad 4 + 2B + C = 6 \quad \Rightarrow \quad 2B + C = 2 \quad \text{(4)} \]

From (2):
\[ -2 + B - C = 5 \quad \Rightarrow \quad B - C = 7 \quad \text{(5)} \]

Now we can solve equations (4) and (5):

From (4):
\[ C = 2 - 2B \quad \text{(6)} \]

Substituting (6) into (5):
\[ B - (2 - 2B) = 7 \quad \Rightarrow \quad B + 2B - 2 = 7 \quad \Rightarrow \quad 3B = 9 \quad \Rightarrow \quad B = 3 \]

Now substituting \(B = 3\) back into (6):
\[ C = 2 - 2(3) = 2 - 6 = -4 \]

### Step 7: Write the Final Partial Fraction Decomposition
Now we have \(A = 2\), \(B = 3\), and \(C = -4\). Thus, we can write:

\[ \frac{6x^2 + 5x - 2}{x(2x + 1)(x - 1)} = \frac{2}{x} + \frac{3}{x - 1} - \frac{4}{2x + 1} \]

### Final Answer:
\[ \frac{6x^2 + 5x - 2}{2x^3 - x^2 - x} = \frac{2}{x} + \frac{3}{x - 1} - \frac{4}{2x + 1} \]

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