Question

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

Answer 1

$\large\underline{\sf{Solution-}}$

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 :-

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$