If `p(x+1)^(2) +q(x^(2)-3x-2)+x+1=0` be an identity in x, then p,q are

Options

A. 2,-2
B. 1,-1
C. 0,0
D. none

Answer

To solve the equation \( p(x+1)^2 + q(x^2 - 3x - 2) + x + 1 = 0 \) as an identity in \( x \), we need to ensure that the coefficients of the polynomial are zero for all powers of \( x \). This means we need to collect like terms and set the coefficients of \( x^2 \), \( x \), and the constant term to zero.

### Step 1: Expand the equation
First, we expand \( p(x+1)^2 \) and \( q(x^2 - 3x - 2) \):

\[ p(x+1)^2 = p(x^2 + 2x + 1) = px^2 + 2px + p \]

\[ q(x^2 - 3x - 2) = qx^2 - 3qx - 2q \]

Now, substituting these back into the original equation:

\[ px^2 + 2px + p + qx^2 - 3qx - 2q + x + 1 = 0 \]

### Step 2: Combine like terms
Next, we combine the terms:

\[ (px^2 + qx^2) + (2px - 3qx + x) + (p - 2q + 1) = 0 \]

This simplifies to:

\[ (p + q)x^2 + (2p - 3q + 1)x + (p - 2q + 1) = 0 \]

### Step 3: Set coefficients to zero
For the equation to be an identity, the coefficients of \( x^2 \), \( x \), and the constant term must all be zero:

1. \( p + q = 0 \) (Coefficient of \( x^2 \))
2. \( 2p - 3q + 1 = 0 \) (Coefficient of \( x \))
3. \( p - 2q + 1 = 0 \) (Constant term)

### Step 4: Solve the equations
From the first equation, we can express \( q \) in terms of \( p \):

\[ q = -p \]

Now substitute \( q = -p \) into the second equation:

\[ 2p - 3(-p) + 1 = 0 \implies 2p + 3p + 1 = 0 \implies 5p + 1 = 0 \implies p = -\frac{1}{5} \]

Now substitute \( p = -\frac{1}{5} \) back into \( q = -p \):

\[ q = -(-\frac{1}{5}) = \frac{1}{5} \]

### Step 5: Final values
Thus, we find:

\[ p = -\frac{1}{5}, \quad q = \frac{1}{5} \]

### Summary
The values of \( p \) and \( q \) are:

\[ p = -\frac{1}{5}, \quad q = \frac{1}{5} \]