If `(x^(4))/((x-1)(x-2))=x^(2)+3x+7+(A)/(x-2)+(B)/(x-1)` then `A=`

Options

A. 7
B. 16
C. -7
D. 9

Answer

To solve the equation

\[ \frac{x^4}{(x-1)(x-2)} = x^2 + 3x + 7 + \frac{A}{x-2} + \frac{B}{x-1} \]

we need to find the value of \( A \).

### Step 1: Rewrite the equation
First, we can rewrite the right-hand side with a common denominator:

\[ x^2 + 3x + 7 + \frac{A}{x-2} + \frac{B}{x-1} = \frac{(x^2 + 3x + 7)(x-2)(x-1) + A(x-1) + B(x-2)}{(x-1)(x-2)} \]

### Step 2: Clear the denominators
Now, we can multiply both sides by \((x-1)(x-2)\) to eliminate the denominators:

\[ x^4 = (x^2 + 3x + 7)(x-2)(x-1) + A(x-1) + B(x-2) \]

### Step 3: Expand the right-hand side
Next, we need to expand the right-hand side. Start with \((x^2 + 3x + 7)(x-2)(x-1)\):

1. First, expand \((x-2)(x-1)\):

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

2. Now, multiply this result by \((x^2 + 3x + 7)\):

\[ (x^2 + 3x + 7)(x^2 - 3x + 2) \]

Expanding this gives:

\[ x^4 + 3x^3 + 7x^2 - 3x^3 - 9x^2 - 21x + 2x^2 + 6x + 14 \]

Combining like terms results in:

\[ x^4 + (3x^3 - 3x^3) + (7x^2 - 9x^2 + 2x^2) + (-21x + 6x) + 14 = x^4 + 0x^3 + 0x^2 - 15x + 14 \]

So we have:

\[ x^4 = x^4 - 15x + 14 + A(x-1) + B(x-2) \]

### Step 4: Substitute values to find A
To find \( A \), we can substitute \( x = 2 \):

\[ 2^4 = 2^4 - 15(2) + 14 + A(2-1) + B(2-2) \]

This simplifies to:

\[ 16 = 16 - 30 + 14 + A(1) + 0 \]

Thus:

\[ 16 = 0 + A \]

So, we find:

\[ A = 16 \]

### Final Answer
The value of \( A \) is:

\[ \boxed{16} \]