If -2 and 3 are the zeroes of the quadratic polynomial x²+(p+1)x+q, t

1.	If -2 and 3 are the zeroes of the quadratic polynomial x²+(p+1)x+q, then find the values of p and q
Answer» $\large\underline{\sf{Given- }}$ -2 and 3 are the zeroes of the quadratic POLYNOMIAL x² + (p + 1) X + q $\large\underline{\sf{To\:Find - }}$ $\: \: \: \: \: \: \: \: \bull \sf \: values \: of \: p \: and \: q$ $\large\underline{\sf{Solution-}}$ We know that $\sf \: If \: \alpha \: and \: \beta \: are \: zeroes \: of \: f(x) = {ax}^{2} + bx + c \: \: then$ $\boxed{\red{\sf Sum\ of\ the\ zeroes \: = \alpha + \beta =\dfrac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$ OR $\boxed{\purple{\tt Sum\ of\ the\ zeroes = \alpha + \beta =\dfrac{-b}{a}}}$ And $\boxed{\red{\sf Product\ of\ the\ zeroes = \alpha \beta =\dfrac{Constant}{coefficient\ of\ x^{2}}}}$ OR $\boxed{\purple{\tt Product\ of\ the\ zeroes = \alpha \beta =\dfrac{c}{a}}}$ Now, Given quadratic polynomial is $\: \: \: \: \: \: \: \: \bull \: \: \sf \: f(x) \: = \: {x}^{2} + (p + 1)x + q$ ↝ On Comparing with ax² + bx + c, we get $\: \: \: \: \: \: \: \: \bull \sf \: a \: = \: 1$ $\: \: \: \: \: \: \: \: \bull \sf \: b \: = \: p + 1$ $\: \: \: \: \: \: \: \: \bull \sf \: c \: = \: q$ ↝ Aɢᴀɪɴ, ↝ -2 and 3 are the zeroes of the quadratic polynomial x² + (p + 1)x + q, $\: \: \: \: \: \: \: \: \bull \sf \: \alpha = - \: 2$ $\: \: \: \: \: \: \: \: \bull \sf \: \beta \: = \: 3$ Now, $\rm :\longmapsto\: \alpha + \beta = - \: \dfrac{b}{a}$ $\rm :\longmapsto\: - 2 + 3 = - \: \dfrac{(p + 1)}{1}$ $\rm :\longmapsto\:1 = - p - 1$ $\bf\implies \:p \: = \: - \: 2$ Also, $\rm :\longmapsto\: \alpha \beta = \dfrac{c}{a}$ $\rm :\longmapsto\:( - 2) \times 3 = \dfrac{q}{1}$ $\bf\implies \:q = - \: 6$ $\overbrace{ \underline { \boxed { \bf \therefore \: The \: value \: of \:p = - 2 \: \: and \: \: q = - 6}}}$