Question

If one of the zeroes of the cubic polynomial ${x^3} + a{x^2} + bx + c$, is -1 then find the product of the other two zeroes.

Answer 1

Hint – In this question let the three roots of the polynomial be $\alpha ,\beta {\text{ and }}\gamma $. $\alpha $ is given as -1, thus using the relation in a cubic polynomial the product of roots is the ratio of negative times the constant term to the coefficient of ${x^3}$, this helps to get the product of the other two roots.

Complete step-by-step answer:

Given cubic polynomial
${x^3} + a{x^2} + bx + c$
As we know that the zeroes are nothing but the roots of the equation.
So, let the roots of this cubic polynomial be $\alpha ,\beta {\text{ and }}\gamma $.
Now it is given that one of the roots of this cubic polynomial is (-1).
So let, $\alpha = - 1$............... (1)
Now as we know that in a cubic polynomial the product of roots is the ratio of negative times the constant term to the coefficient of $x^3$.
$ \Rightarrow \alpha \beta \lambda = \dfrac{{ - {\text{constant term}}}}{{{\text{coefficient of }}{x^3}}}$
$ \Rightarrow \alpha \beta \lambda = \dfrac{{ - c}}{1} = - c$
Now from equation (1) we have,
$ \Rightarrow \left( { - 1} \right)\beta \lambda = - c$
Now as we see negative sign is cancel out from both sides so we have,
$ \Rightarrow \beta \lambda = c$
So the product of the other two roots or zeros is c.
So this is the required answer.

Note – In this question the tricky part was that we don’t actually need to find the all three roots of the cubic polynomial to get the product of the roots, thus remembering this direct relations between sum and the product of roots always helps in saving a lot of time. The sum of roots $\alpha + \beta + \gamma = \dfrac{{ - b}}{a}$ for any cubic polynomial of the form ${x^3} + a{x^2} + bx + c$.