If the zeroes of the quadratic polynomial `x^(2)+(a+1)x+b` are 2 and -3 ,then

Options

A. `x^(2)+(a+1)x+b` are 2 and -3 ,then
B. `a=5,b=-1`
C. `a=2,b=-6`
D. `a=0,b=-6`

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) in the quadratic polynomial \( x^2 + (a+1)x + b \) given that its zeroes are \( 2 \) and \( -3 \).

### Step-by-Step Solution:

1. **Identify the Zeroes**:
The zeroes of the polynomial are given as \( \alpha = 2 \) and \( \beta = -3 \).

2. **Use the Sum of Zeroes**:
The sum of the zeroes \( \alpha + \beta \) can be calculated:
\[ \alpha + \beta = 2 + (-3) = -1 \]
According to the relationship between the coefficients and the roots of a polynomial, we have:
\[ \alpha + \beta = -\frac{b}{a} \]
Here, \( a = 1 \) (the coefficient of \( x^2 \)) and \( b = a + 1 \) (the coefficient of \( x \)). Therefore:
\[ -1 = -\frac{a + 1}{1} \]
Simplifying this gives:
\[ -1 = -(a + 1) \]
Thus:
\[ 1 = a + 1 \]
Rearranging gives:
\[ a = 0 \]

3. **Use the Product of Zeroes**:
The product of the zeroes \( \alpha \cdot \beta \) can be calculated:
\[ \alpha \cdot \beta = 2 \cdot (-3) = -6 \]
According to the relationship between the coefficients and the roots of a polynomial, we have:
\[ \alpha \cdot \beta = \frac{c}{a} \]
Here, \( c = b \) (the constant term). Therefore:
\[ -6 = \frac{b}{1} \]
Thus:
\[ b = -6 \]

4. **Final Values**:
We have found:
\[ a = 0 \quad \text{and} \quad b = -6 \]

### Conclusion:
The values of \( a \) and \( b \) are \( 0 \) and \( -6 \) respectively.