Question

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 will be

• A. $\frac{-c}{a}$
• B. $\frac{c}{a}$
• C. 0.0
• D. $\frac{-b}{a}$

Answer 1

The correct option is B $\frac{c}{a}$

Let $\alpha ,\beta and\gamma$ be the three zeroes 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 $(let\alpha =0)$.

We know that sum of the product of zeroes taken two at a time = $\frac{c}{a}$
$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$
$\Rightarrow 0+\beta \gamma +\gamma \times 0=\frac{c}{a}$
$\Rightarrow 0+\beta \gamma +0=\frac{c}{a}$
$\Rightarrow \beta \gamma =\frac{c}{a}$

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