Question

Suppose $\alpha ,\beta .\gamma$ are roots of $x^3+x^2+2x+3=0$. If $f\left(x\right)=0$ is a cubic polynomial equation whose roots are $\alpha +\beta ,\beta +\gamma ,\gamma +\alpha$ then $f\left(x\right)=$

• A. $x^3+2x^2-3x-1$
• B. $x^3+2x^2-3x+1$
• C. $x^3+2x^2+3x-1$
• D. $x^3+2x^2+3x+1$

Answer 1

The correct option is C $x^3+2x^2+3x-1$

Given that $\alpha ,\beta ,\gamma$ are the roots of $x^3+x^2+2x+3=0$

so we get $\alpha +\beta +\gamma =-1,\alpha \beta +\beta \gamma +\gamma \alpha =2,\alpha \beta \gamma =-3$

Given that $\alpha +\beta ,\beta +\gamma ,\gamma +\alpha$ are the roots of $f\left(x\right)=0$

let $f\left(x\right)=x^3+b{x}^2+cx+d=0$

$-b=2(\alpha +\beta +\gamma )=-2$

$\Rightarrow b=2$

$\Rightarrow c=(\alpha +\beta )(\beta +\gamma )+(\alpha +\gamma )(\beta +\gamma )+(\alpha +\beta )(\alpha +\gamma )={\alpha }^2+{\beta }^2+{\gamma }^2+3(\alpha \beta +\beta \gamma +\alpha \gamma )$

$\Rightarrow c={(\alpha +\beta +\gamma )}^2+\alpha \beta +\beta \gamma +\alpha \gamma =3$

$\Rightarrow -d=(\alpha +\beta )(\beta +\gamma )(\gamma +\alpha )=(\alpha +\beta +\gamma )(\alpha \beta +\beta \gamma +\gamma \alpha )-\alpha \beta \gamma =(-1)\left(2\right)-(-3)=1$

$\Rightarrow d=-1$

Therefore $f\left(x\right)=x^3+2x^2+3x-1$