Question

If one of the zeroes of the cubic polynomial x3 + ax² + bx + c is -1, then the product of the
other two zeroes is​

Answer 1

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

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,

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

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

$\boxed{ \rm{ \alpha \beta = \: \dfrac{c}{a}}}$