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 is

(a) $\frac{-c}{a}$ (b) $\frac{c}{a}$ (c) 0 (d) $\frac{-b}{a}$

Answer 1

Let $p\left(x\right)=a{x}^3+b{x}^2+cx+d$.
Now 0 is the zero of the polynomial.
So, p(0) = 0.
$$\begin{gathered} \Rightarrow a{\left(0\right)}^3+b{\left(0\right)}^2+c\left(0\right)+d=0 \\ \Rightarrow d=0 \end{gathered}$$
So,
$p\left(x\right)=a{x}^3+b{x}^2+cx=x\left(a{x}^2+bx+c\right)$
Putting p(x) = 0, we get
x = 0 or $a{x}^2+bx+c=0$ .....(1)
Let α, β be the other zeroes of $a{x}^2+bx+c=0$.
So, $\alpha \beta =\frac{c}{a}$

Hence, the correct answer is option B.