# Choose the correct answer from the given four options in the following questions:Given that one of the zeroes of the cubic polynomial $a x^{3}+b x^{2}+c x+d$ is zero, the product of the other two zeroes is(A) $-\frac{c}{a}$(B) $\frac{c}{a}$(C) 0(D) $-\frac{b}{a}$

Given:

One of the zeroes of the cubic polynomial $a x^{3}+b x^{2}+c x+d$ is zero.

To do:

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

Solution:

Let $p(x)=a x^{3}+b x^{2}+c x+d$

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

Let $\alpha, \beta$ and $\gamma$ are the zeroes of the cubic polynomial $p(x)$, where $\alpha=0$.

We know that,
Sum of the product of two zeroes taken at a time $=\frac{c}{a}$

$\Rightarrow \alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a}$

$\Rightarrow 0 \times \beta+\beta \gamma+\gamma \times 0=\frac{c}{a}$

$\Rightarrow 0+\beta \gamma+0=\frac{c}{a}$

$\Rightarrow \beta \gamma=\frac{c}{a}$
Hence, the product of other two zeroes is $\frac{c}{a}$.

Tutorialspoint

Tutorialspoint Tag Line