If `(x^(3)-x^(2)-13x+25)/((x-2)^(2)(x-3)^2)= (A)/(x-2)+(B)/((x-2)^2)+(C )/((x-3))+(D)/((x-3)^2)` then A, B, C, D are

Options

A. 1,3,0,4
B. 1,0,3,4
C. 1,4,0,3
D. none

Answer

To solve the equation

\[ \frac{x^3 - x^2 - 13x + 25}{(x-2)^2 (x-3)^2} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{(x-3)} + \frac{D}{(x-3)^2} \]

we will find the values of \(A\), \(B\), \(C\), and \(D\) through the method of partial fractions.

### Step 1: Set Up the Equation

We start by expressing the left-hand side in terms of the right-hand side:

\[ x^3 - x^2 - 13x + 25 = A(x-2)(x-3)^2 + B(x-3)^2 + C(x-2)^2(x-3) + D(x-2)^2 \]

### Step 2: Expand the Right-Hand Side

We need to expand the right-hand side to compare coefficients.

1. **Expand \(A(x-2)(x-3)^2\)**:
\[ = A(x-2)(x^2 - 6x + 9) = A(x^3 - 8x^2 + 21x - 18) \]

2. **Expand \(B(x-3)^2\)**:
\[ = B(x^2 - 6x + 9) \]

3. **Expand \(C(x-2)^2(x-3)\)**:
\[ = C((x^2 - 4x + 4)(x-3)) = C(x^3 - 7x^2 + 12x - 12) \]

4. **Expand \(D(x-2)^2\)**:
\[ = D(x^2 - 4x + 4) \]

### Step 3: Combine All Terms

Now combine all the expanded terms:

\[ (A + C)x^3 + (-8A - 7C + B + D)x^2 + (21A - 6B + 12C - 4D)x + (-18A + 9B - 12C + 4D) = x^3 - x^2 - 13x + 25 \]

### Step 4: Set Up the System of Equations

By equating the coefficients from both sides, we get the following system of equations:

1. \(A + C = 1\) (coefficient of \(x^3\))
2. \(-8A - 7C + B + D = -1\) (coefficient of \(x^2\))
3. \(21A - 6B + 12C - 4D = -13\) (coefficient of \(x\))
4. \(-18A + 9B - 12C + 4D = 25\) (constant term)

### Step 5: Solve the System of Equations

1. From equation (1):
\[ C = 1 - A \]

2. Substitute \(C\) into equations (2), (3), and (4):

- For equation (2):
\[ -8A - 7(1 - A) + B + D = -1 \implies -8A - 7 + 7A + B + D = -1 \implies -A + B + D = 6 \quad \text{(Equation 5)} \]

- For equation (3):
\[ 21A - 6B + 12(1 - A) - 4D = -13 \implies 21A - 6B + 12 - 12A - 4D = -13 \implies 9A - 6B - 4D = -25 \quad \text{(Equation 6)} \]

- For equation (4):
\[ -18A + 9B - 12(1 - A) + 4D = 25 \implies -18A + 9B - 12 + 12A + 4D = 25 \implies -6A + 9B + 4D = 37 \quad \text{(Equation 7)} \]

### Step 6: Solve Equations 5, 6, and 7

Now, we have three equations (5, 6, and 7) to solve for \(A\), \(B\), and \(D\).

1. From equation (5):
\[ B + D = A + 6 \]

2. Substitute \(B\) from equation (5) into equations (6) and (7) and solve for \(A\), \(B\), \(C\), and \(D\).

After solving the equations, we find:

- \(A = 1\)
- \(B = 3\)
- \(C = 0\)
- \(D = 4\)

### Final Values

Thus, the values are:
\[ A = 1, \quad B = 3, \quad C = 0, \quad D = 4 \]