Resolve the following into partial fractions.
`(2x-1)/((1-x-x^(2))(x+2))`

Answer

To resolve the expression \(\frac{2x-1}{(1-x-x^2)(x+2)}\) into partial fractions, we will follow these steps:

### Step 1: Set up the partial fraction decomposition
We express \(\frac{2x-1}{(1-x-x^2)(x+2)}\) as a sum of partial fractions. Since \(1-x-x^2\) is a quadratic polynomial, we can write:

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

where \(A\), \(B\), and \(C\) are constants to be determined.

### Step 2: Combine the fractions
To combine the fractions on the right side, we find a common denominator:

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

### Step 3: Eliminate the denominators
Since the denominators are the same, we can equate the numerators:

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

### Step 4: Expand the right side
Now we expand the right-hand side:

\[ A(1 - x - x^2) = A - Ax - Ax^2 \]
\[ (Bx + C)(x + 2) = Bx^2 + 2Bx + Cx + 2C = (B)x^2 + (2B + C)x + 2C \]

Combining these gives:

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

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

1. Coefficient of \(x^2\): \(B - A = 0\) (1)
2. Coefficient of \(x\): \(2B + C - A = 2\) (2)
3. Constant term: \(A + 2C = -1\) (3)

### Step 6: Solve the system of equations
From equation (1), we have:

\[ B = A \]

Substituting \(B = A\) into equation (2):

\[ 2A + C - A = 2 \implies A + C = 2 \quad (4) \]

Substituting \(B = A\) into equation (3):

\[ A + 2C = -1 \quad (5) \]

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

From (4): \(C = 2 - A\)

Substituting into (5):

\[ A + 2(2 - A) = -1 \]
\[ A + 4 - 2A = -1 \]
\[ -A + 4 = -1 \]
\[ -A = -5 \implies A = 5 \]

Now substituting \(A = 5\) back into (4):

\[ 5 + C = 2 \implies C = 2 - 5 = -3 \]

And since \(B = A\):

\[ B = 5 \]

### Step 7: Write the final partial fractions
Now we can substitute \(A\), \(B\), and \(C\) back into the partial fraction decomposition:

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

### Final Answer
Thus, the partial fraction decomposition is:

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