If `(x^(2)+x+1)/(x^(2)+2x+1)=A+B/(x+1)+C/(x+1)^(2)` then A - B =

Options

A. 4C
B. 2C
C. 3C
D. `4C+1`

Answer

To solve the equation \(\frac{x^2 + x + 1}{x^2 + 2x + 1} = A + \frac{B}{x+1} + \frac{C}{(x+1)^2}\), we will follow these steps:

### Step 1: Rewrite the right-hand side with a common denominator
The right-hand side can be expressed with a common denominator of \((x + 1)^2\):
\[ A + \frac{B}{x+1} + \frac{C}{(x+1)^2} = \frac{A(x + 1)^2 + B(x + 1) + C}{(x + 1)^2} \]

### Step 2: Set the left-hand side equal to the right-hand side
Now we can equate the left-hand side to the right-hand side:
\[ \frac{x^2 + x + 1}{(x + 1)^2} = \frac{A(x + 1)^2 + B(x + 1) + C}{(x + 1)^2} \]
Since the denominators are the same, we can equate the numerators:
\[ x^2 + x + 1 = A(x + 1)^2 + B(x + 1) + C \]

### Step 3: Expand the right-hand side
Now, we will expand the right-hand side:
\[ A(x + 1)^2 = A(x^2 + 2x + 1) = Ax^2 + 2Ax + A \]
\[ B(x + 1) = Bx + B \]
Combining these, we have:
\[ Ax^2 + (2A + B)x + (A + B + C) \]

### Step 4: Equate coefficients
Now we can equate the coefficients from both sides of the equation:
1. Coefficient of \(x^2\): \(A = 1\)
2. Coefficient of \(x\): \(2A + B = 1\)
3. Constant term: \(A + B + C = 1\)

### Step 5: Solve for \(A\), \(B\), and \(C\)
From the first equation, we have:
\[ A = 1 \]
Substituting \(A = 1\) into the second equation:
\[ 2(1) + B = 1 \implies 2 + B = 1 \implies B = 1 - 2 = -1 \]
Now substituting \(A = 1\) and \(B = -1\) into the third equation:
\[ 1 - 1 + C = 1 \implies C = 1 \]

### Step 6: Find \(A - B\)
Now that we have \(A\) and \(B\):
\[ A - B = 1 - (-1) = 1 + 1 = 2 \]

### Final Answer
Thus, \(A - B = 2\).

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