If one of the zeroes of the cubic polynomial x3 + ax² + bx + c is -1

1.	If one of the zeroes of the cubic polynomial x3 + ax² + bx + c is -1, then the product of theother two zeroes is
Answer» $\large\underline{\sf{Solution-}}$ Let assume that $\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: {x}^{3} +a{x}^{2} + bx + c}$ Then further given that $\rm :\longmapsto\: \alpha = - 1$ Since, - 1 the zero of the POLYNOMIAL, it means $\rm :\longmapsto\: {( - 1)}^{3} + a {( - 1)}^{2} + b( - 1) + c = 0$ $\rm :\longmapsto\: - 1+ a - b+ c = 0$ $\bf :\longmapsto\: c = - a + b + 1$ Now, we know that, $\boxed{\red{\sf Product\ of\ the\ zeroes= - \: \frac{Constant}{coefficient\ of\ x^{3}}}}$ $\rm :\implies\: \alpha \beta \gamma = - \: \dfrac{c}{1}$ $\rm :\longmapsto\:( - 1) \beta \gamma = - c$ $\rm :\longmapsto\: \beta \gamma = c$ $\rm :\longmapsto\: \beta \gamma = - a + b + 1$ ADDITIONAL INFORMATION :- For CUBIC polynomial $\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$ $\boxed{ \rm{ \alpha + \beta + \gamma = - \: \dfrac{b}{a} }}$ $\boxed{ \rm{ \alpha \beta + \beta \gamma + \gamma \alpha = \: \dfrac{c}{a} }}$ $\boxed{ \rm{ \alpha \beta \gamma = - \: \dfrac{d}{a} }}$ For quadratic polynomial, $\red{\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: a {x}^{2} + b {x} + c, \: then}$ $\boxed{ \rm{ \alpha + \beta = - \: \dfrac{b}{a}}}$ $\boxed{ \rm{ \alpha \beta = \: \dfrac{c}{a}}}$