If the zeroes of the quadratic polynomial \(x^2 + (a+1)x + b\) are \(2\) and \(-3\), then
\(a = -7,\; b = -1\)
\(a = 5,\; b = -1\)
\(a = 2,\; b = -6\)
\(a = 0,\; b = -6\)
Step 1: Recall the relationships of zeroes with coefficients.
If \(\alpha\) and \(\beta\) are the zeroes of \(x^2 + px + q\), then
\(\alpha + \beta = -p\)
\(\alpha \cdot \beta = q\)
Step 2: Apply to our polynomial.
Here, polynomial is \(x^2 + (a+1)x + b\).
So, coefficient of \(x\) is \(a+1\) and constant term is \(b\).
Thus,
Sum of zeroes \(= - (a+1)\)
Product of zeroes \(= b\)
Step 3: Use the given zeroes.
The zeroes are \(2\) and \(-3\).
So,
\(\alpha + \beta = 2 + (-3) = -1\)
\(\alpha \cdot \beta = 2 \times (-3) = -6\)
Step 4: Form equations for \(a\) and \(b\).
From sum: \(- (a+1) = -1\)
\(a+1 = 1\)
\(a = 0\)
From product: \(b = -6\)
Step 5: Final answer.
So the correct values are \(a = 0\) and \(b = -6\).
This corresponds to option D.