Choose the correct answer from the given four options in the following questions:
If one of the zeroes of the cubic polynomial $ x^{3}+a x^{2}+b x+c $ is $ -1 $, then the product of the other two zeroes is
(A) $ b-a+1 $
(B) $ b-a-1 $
(C) $ a-b+1 $
(D) $ a-b-1 $

Given: 

One of the zeroes of the cubic polynomial \( x^{3}+a x^{2}+b x+c \) is \( -1 \).

To do: 

We have to find the product of the other two zeroes.

Solution:

Let $p(x)=x^{3}+ax^{2}+bx+c$

One of the zeroes of the cubic polynomial $p(x)$ is $-1$.

Let $\alpha, \beta$ and $\gamma$ be the zeroes of the cubic polynomial $p(x)$, where $\alpha=-1$.

$p(-1)=0$

$\Rightarrow (-1)^{3}+a(-1)^{2}+b(-1)+c=0$

$\Rightarrow -1+a-b+c=0$

$\Rightarrow c=1-a+b$.............(i)

We know that,

Product of all three zeroes $=-\frac{\text { Constant term }}{\text { Coefficient of } x^{3}}$

$=-\frac{c}{1}$

$=-c$

$\alpha \beta \gamma=-c$

$\Rightarrow (-1) \beta \gamma=-c$

$\Rightarrow \beta \gamma=c$

$\Rightarrow \beta \gamma=1-a+b$           [From (i)]

Hence, the product of the other two roots is $1-a+b$.

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Updated on: 10-Oct-2022

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