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

Options

A. `a =- 7, b =- 1`
B. `a = 5, b=-1`
C. `a =2, b =- 6`
D. `a =0, b=-6`

Explanation

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

### Step-by-Step Solution:

1. **Use the fact that the sum of the zeroes of a quadratic polynomial \(ax^2 + bx + c\) is \(-\frac{b}{a}\):**

Given zeroes are 2 and -3.
\[ \text{Sum of the zeroes} = 2 + (-3) = -1 \]
For the polynomial \(x^2 + (a+1)x + b\), the sum of the zeroes is:
\[ -(a+1) \]
Equate the two expressions:
\[ -(a+1) = -1 \]
Simplify to find \(a\):
\[ a + 1 = 1 \]
\[ a = 0 \]

2. **Use the fact that the product of the zeroes of a quadratic polynomial \(ax^2 + bx + c\) is \(\frac{c}{a}\):**

Given zeroes are 2 and -3.
\[ \text{Product of the zeroes} = 2 \times (-3) = -6 \]
For the polynomial \(x^2 + (a+1)x + b\), the product of the zeroes is:
\[ b \]
Equate the two expressions:
\[ b = -6 \]

3. **Verify the values of \(a\) and \(b\):**

Substitute \(a = 0\) and \(b = -6\) back into the polynomial:
\[ x^2 + (0+1)x + (-6) = x^2 + x - 6 \]
Check the zeroes of \(x^2 + x - 6\):
\[ x^2 + x - 6 = (x - 2)(x + 3) \]
The zeroes are indeed 2 and -3, confirming our solution.

### Final Answer:
\[ a = 0, \quad b = -6 \]