Question

Question 5
Given that one of the zeroes of the cubic polynomial $a{x}^3+b{x}^2+cx+d$ is zero, the product of the other two zeroes is
(a)$\frac{-c}{a}$
(b)$\frac{c}{a}$
(c)0
(d)$\frac{-b}{a}$

Answer 1

Let $\alpha ,\beta and\gamma$ are the zeros of $p\left(x\right)=a{x}^3+b{x}^2+cx+d$
Given that, one of the zeroes of the cubic polynomial p(x) is zero $(\alpha =0)$.
We know that,
Sum of product of zeroes taken two at a time = $\frac{c}{a}$
$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$
$\Rightarrow 0\times \beta \gamma +\gamma \times 0=\frac{c}{a}$
$\Rightarrow 0+\beta \gamma +0=\frac{c}{a}$
$\Rightarrow \beta \gamma =\frac{c}{a}$
Hence, product of other two zeroes = $\frac{c}{a}$
So option B is correct.