If -6 and 3 are zeroes of the quadratic polynomial x² + (a+1)x + b, t

1.	If -6 and 3 are zeroes of the quadratic polynomial x² + (a+1)x + b, then a and b are:-
Answer» $\large\underline{\sf{Solution-}}$ GIVEN that, $\red{\rm :\longmapsto\: - 6 \: and \: 3 \: are \: zeroes \: of \: {x}^{2} + (a + 1)x + b}$ We KNOW that, $\boxed{ \purple{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$ $\rm :\implies\: - 6 \times 3 = \dfrac{b}{1}$ $\bf\implies \:b = - 18$ ALSO, $\boxed{\purple{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$ $\rm :\implies\: - 6 + 3 = - \: \dfrac{(a + 1)}{1}$ $\rm :\longmapsto\: - 3 = - a - 1$ $\rm :\longmapsto\: a = 3 - 1$ $\bf\implies \:a = 2$ ADDITIONAL INFORMATION :- $\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$ $\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$ $\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$ $\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$