# 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) $a-b-1$

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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 $1-a+b$.

Updated on 10-Oct-2022 13:27:08

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