Question

If $\alpha ,\beta ,\gamma$ are the zeroes of a cubic polynomial $a{x}^3+b{x}^2+cx+d$, then the sum of zeroes is equal to

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

Answer 1

The correct option is A $\frac{-b}{a}$

Given that

$p\left(x\right)=a{x}^3+b{x}^2+cx+d$ .........(1) where $a\ne 0$ is a cubic polynomial

And $\alpha ,\beta ,\gamma$ are the zeroes of the polynomial $p\left(x\right)$

We can also write the polynomial in this way

$p\left(x\right)=(x-\alpha )(x-\beta )(x-\gamma )$

$\Rightarrow p\left(x\right)=x^3-(\alpha +\beta +\gamma )x^2+(\alpha \beta +\beta \gamma +\gamma \alpha )x-\alpha \beta \gamma$ .............(2)

Equations (1) and (2) are same

Comparing the coefficients

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

On solving we get the following results

$\alpha +\beta +\gamma =\frac{-b}{a}=$ Sum of the roots

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$

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

Hence, option $A$ is correct.