If (x+1) is the HCF of `(a x^(2) + bx + c)` and `(bx^(2) + ax + c)`, then the value of c is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the value of \( c \) given that \( (x+1) \) is the HCF of \( (a x^2 + b x + c) \) and \( (b x^2 + a x + c) \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: Since \( (x+1) \) is the HCF, it means that both polynomials \( (a x^2 + b x + c) \) and \( (b x^2 + a x + c) \) must be divisible by \( (x+1) \). Therefore, substituting \( x = -1 \) into both polynomials should yield zero.

2. **Substituting into the First Polynomial**:
\[ a(-1)^2 + b(-1) + c = 0 \]
Simplifying this gives:
\[ a - b + c = 0 \quad \text{(Equation 1)} \]

3. **Substituting into the Second Polynomial**:
\[ b(-1)^2 + a(-1) + c = 0 \]
Simplifying this gives:
\[ b - a + c = 0 \quad \text{(Equation 2)} \]

4. **Setting Up the Equations**: Now we have two equations:
- From Equation 1: \( c = b - a \)
- From Equation 2: \( c = a - b \)

5. **Equating the Two Expressions for \( c \)**:
\[ b - a = a - b \]
Rearranging gives:
\[ b + b = a + a \implies 2b = 2a \implies b = a \]

6. **Substituting \( b = a \) back into either equation**:
Using Equation 1:
\[ c = b - a = a - a = 0 \]

7. **Conclusion**: Therefore, the value of \( c \) is \( 0 \).

### Final Answer:
\[ c = 0 \]