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

Question

Accepted Answer

To determine why Option 2 is the correct answer, we need to analyze the given quadratic polynomial and its zeroes. The polynomial is $x^2 + (a + 1)x + b$, and we know that $-2$ and $3$ are its zeroes.

### Explanation for Option 2:

1. **Sum of Zeroes**: According to Vieta's formulas, the sum of the zeroes of a quadratic polynomial $ax^2 + bx + c$ is given by $-b/a$. For the polynomial $x^2 + (a + 1)x + b$, the sum of the zeroes $-2$ and $3$ is:
   $$
   -2 + 3 = 1
   $$
   This should equal $-(a + 1)$. Therefore, we have:
   $$
   -(a + 1) = 1 \implies a + 1 = -1 \implies a = -2
   $$

2. **Product of Zeroes**: The product of the zeroes is given by $c/a$. For the polynomial $x^2 + (a + 1)x + b$, the product of the zeroes $-2$ and $3$ is:
   $$
   (-2) \times 3 = -6
   $$
   This should equal $b$. Therefore, we have:
   $$
   b = -6
   $$

Thus, the values $a = -2$ and $b = -6$ satisfy both the sum and product of zeroes conditions, confirming that Option 2 is correct.

### Explanation for Why Other Options are Incorrect:

1. **Option 1 (a = 2; b = 6)**:
   - If $a = 2$, then the sum of the zeroes would be $-(2 + 1) = -3$, which does not match the actual sum of $1$.
   - If $b = 6$, then the product of the zeroes would be $6$, which does not match the actual product of $-6$.

2. **Option 3 (a = -2; b = 6)**:
   - While $a = -2$ correctly satisfies the sum of zeroes, $b = 6$ does not satisfy the product of zeroes, which should be $-6$.

3. **Option 4 (a = 2; b = -6)**:
   - If $a = 2$, then the sum of the zeroes would be $-(2 + 1) = -3$, which does not match the actual sum of $1$.
   - Although $b = -6$ correctly satisfies the product of zeroes, the incorrect value of $a$ makes this option invalid.

In conclusion, only Option 2 satisfies both the sum and product of zeroes conditions, making it the correct choice.