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 $\frac{c}{a}$.

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