If the zeroes of the quadratic polynomial x^{ | vedclass

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(A) Let $p(x) = x^{2} + (a+1)x + b$.Given that $2$ and $-3$ are the zeroes of the quadratic polynomial $p(x)$.Therefore,$p(2) = 0$ and $p(-3) = 0$.For

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$a=0, b=-6$

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(A) Let $p(x) = x^{2} + (a+1)x + b$.Given that $2$ and $-3$ are the zeroes of the quadratic polynomial $p(x)$.Therefore,$p(2) = 0$ and $p(-3) = 0$.For $x = 2$:$2^{2} + (a+1)(2) + b = 0$$4 + 2a + 2 + b = 0$$2a + b = -6$ ..... $(i)$For $x = -3$:$(-3)^{2} + (a+1)(-3) + b = 0$$9 - 3a - 3 + b = 0$$-3a + b = -6$ ..... $(ii)$Subtracting equation $(ii)$ from equation $(i)$:$(2a + b) - (-3a + b) = -6 - (-6)$$2a + b + 3a - b = 0$$5a = 0 \Rightarrow a = 0$.Substituting $a = 0$ in equation $(i)$:$2(0) + b = -6 \Rightarrow b = -6$.Thus,the required values are $a = 0$ and $b = -6$.

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

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