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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