Question

α, β and γ are the zeroes of cubic polynomial $p\left(x\right)=a{x}^3+b{x}^2+cx+d$, $(a\ne 0)$. Then product of their zeroes [αβγ] is?

Answer 1

We have $p\left(x\right)=a{x}^3+b{x}^2+cx+d$, $(a\ne 0)$ $.......\left[1\right]$

If α, β and γ are the zeroes of the cubic polynomial P(x).

then,

$$\begin{gathered} p\left(x\right)=(x-\alpha )(x-\beta )(x-\gamma ) \\ \Rightarrow p\left(x\right)=x^3-(\alpha +\beta +\gamma )x^2+(\alpha \beta +\beta \gamma +\gamma \alpha )x-\left(\alpha \beta \gamma \right).......\left[2\right] \end{gathered}$$

Since, both equation $\left[1\right]$ and $\left[2\right]$ are same.

So, by comparing the coefficients, we get

$\frac{a}{1}=\frac{b}{-\alpha -\beta -\gamma }=\frac{c}{\alpha \beta +\beta \gamma +\gamma \alpha }=\frac{d}{-\alpha \beta \gamma }$

On solving, we get

$\alpha +\beta +\gamma =\frac{-b}{a}$ $=$ sum of roots.

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$$=$sum of product of the roots.

$\alpha \beta \gamma =\frac{-d}{a}=$product of the roots.

Hence, the product of zeroes is $\mathit{\alpha }\mathit{\beta }\mathit{\gamma }\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{-}\mathbf{d}}{\mathbf{a}}$.