Let `f(x)=x^(2)+ax+b` be a quadratic polynomial in which a and b are integers. If for a given integer `n, f(n) f(n+1)=f(m)` for some integer m, then the value of m is

Options

A. `n(a+b)+ab`
B. `n^(2)+an+b`
C. `n(n+1)+an+b`
D. `n^(2)n+a+b`

Answer

To solve the problem, we start with the quadratic polynomial given by \( f(x) = x^2 + ax + b \), where \( a \) and \( b \) are integers. We need to find the value of \( m \) such that \( f(n) \cdot f(n+1) = f(m) \) for some integer \( m \).

### Step-by-Step Solution:

1. **Calculate \( f(n) \)**:
\[ f(n) = n^2 + an + b \]

2. **Calculate \( f(n+1) \)**:
\[ f(n+1) = (n+1)^2 + a(n+1) + b = n^2 + 2n + 1 + an + a + b = n^2 + (a + 2)n + (b + a + 1) \]

3. **Calculate the product \( f(n) \cdot f(n+1) \)**:
\[ f(n) \cdot f(n+1) = (n^2 + an + b) \cdot (n^2 + (a + 2)n + (b + a + 1)) \]
This product can be expanded, but we will focus on the key terms that will help us find \( m \).

4. **Use the property of roots**:
The roots \( \alpha \) and \( \beta \) of the polynomial can be expressed as:
\[ \alpha + \beta = -a \quad \text{and} \quad \alpha \beta = b \]
Thus, we can express \( f(n) \) and \( f(n+1) \) in terms of \( \alpha \) and \( \beta \):
\[ f(n) = (n - \alpha)(n - \beta) \]
\[ f(n+1) = ((n+1) - \alpha)((n+1) - \beta) \]

5. **Set up the equation**:
From the problem statement, we have:
\[ f(n) \cdot f(n+1) = f(m) \]
This implies:
\[ (n - \alpha)(n - \beta)((n + 1) - \alpha)((n + 1) - \beta) = (m - \alpha)(m - \beta) \]

6. **Simplify the expression**:
The left-hand side can be simplified to:
\[ (n^2 + an + b)((n^2 + (a + 2)n + (b + a + 1))) \]
After simplification, we find that the expression can be set equal to \( f(m) \).

7. **Find \( m \)**:
By equating the expressions, we can derive:
\[ m = n(n + 1) + an + b \]

8. **Final Result**:
Thus, the value of \( m \) is:
\[ m = n(n + 1) + an + b \]