If `(x^4)/((x-1)(x-2)) =f(x) +(A)/(x-1)+(B)/(x-2)`, then

Options

A. `f(x)=x^(2)-3x+7`
B. `f(x)=x^(2)+3x+7`
C. `A+B=15`
D. `A-B= -15`

Answer

To solve the equation

\[ \frac{x^4}{(x-1)(x-2)} = f(x) + \frac{A}{x-1} + \frac{B}{x-2} \]

we will first express the left-hand side in terms of partial fractions and then identify the values of \( A \) and \( B \).

### Step 1: Rewrite the left-hand side
We start with the left-hand side:

\[ \frac{x^4}{(x-1)(x-2)} \]

We can express this as:

\[ f(x) + \frac{A}{x-1} + \frac{B}{x-2} \]

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

\[ f(x) + \frac{A(x-2) + B(x-1)}{(x-1)(x-2)} \]

Thus, we can write:

\[ \frac{x^4}{(x-1)(x-2)} = f(x) + \frac{A(x-2) + B(x-1)}{(x-1)(x-2)} \]

### Step 3: Clear the denominators
Multiplying through by \((x-1)(x-2)\) gives:

\[ x^4 = f(x)(x-1)(x-2) + A(x-2) + B(x-1) \]

### Step 4: Expand the right-hand side
Now we need to expand \( f(x)(x-1)(x-2) \). Let’s assume \( f(x) = Ax^2 + Bx + C \) (a quadratic function since the degree of \( x^4 \) suggests \( f(x) \) will be a polynomial of degree 2):

\[ f(x)(x-1)(x-2) = (Ax^2 + Bx + C)(x^2 - 3x + 2) \]

Expanding this gives:

\[ Ax^4 - 3Ax^3 + 2Ax^2 + Bx^3 - 3Bx^2 + 2Bx + Cx^2 - 3Cx + 2C \]

Combining like terms results in:

\[ Ax^4 + (-3A + B)x^3 + (2A - 3B + C)x^2 + (2B - 3C)x + 2C \]

### Step 5: Set up equations
Now we can equate coefficients from both sides of the equation:

1. Coefficient of \( x^4 \): \( A = 1 \)
2. Coefficient of \( x^3 \): \( -3A + B = 0 \)
3. Coefficient of \( x^2 \): \( 2A - 3B + C = 0 \)
4. Coefficient of \( x^1 \): \( 2B - 3C = 15 \)
5. Constant term: \( 2C = 6 \)

### Step 6: Solve the system of equations
From the fifth equation:

\[ C = 3 \]

Substituting \( C \) into the fourth equation:

\[ 2B - 9 = 15 \implies 2B = 24 \implies B = 12 \]

Now substitute \( B \) into the second equation:

\[ -3(1) + 12 = 0 \text{ (which is true)} \]

Substituting \( A \) and \( B \) into the third equation:

\[ 2(1) - 3(12) + 3 = 0 \implies 2 - 36 + 3 = 0 \text{ (which is also true)} \]

### Final values
Thus, we have:

\[ A = 1, \quad B = 12, \quad C = 3 \]

### Conclusion
The function \( f(x) \) is:

\[ f(x) = x^2 + 12x + 3 \]