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

Answer: (D)

$x^3+2x^2+3x+1$

Roots of the given Equation $x^3+x^2+2x+3=0$ are $\alpha ,\beta ,\gamma$

Now ,

$\alpha +\beta +\gamma =-1$

$\sum \alpha \beta =2$

$\alpha \beta \gamma =-3$

Now, defining a equation having roots $\alpha +\beta ,\beta +\gamma ,\gamma +\alpha$ as $x^3+a{x}^2+bx+c=0$

So,

$a$=sum of roots=$2(\alpha +\beta +\gamma )$

$a=-2$

$b$ = Sum of the roots two at a time = $(\alpha +\beta )(\gamma +\alpha )+(\gamma +\alpha )(\beta +\gamma )+(\beta +\gamma )(\alpha +\beta )$

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

$b=(\alpha +\beta +\gamma {)}^2+(\alpha \beta +\gamma \alpha +\gamma \beta )$

$b=(-1{)}^2+2$

$b=3$

$c$ = Product of roots = $(\alpha +\beta )(\alpha +\gamma )(\gamma +\beta )$

$c=\left(\right(\alpha +\beta +\gamma )-\gamma )\left)\right((\alpha +\beta +\gamma )-\beta \left)\right((\alpha +\beta +\gamma )-\alpha )$

this expression is similar to the equation$(x-\alpha )(x-\beta )(x-\gamma )$ as $x=(\alpha +\beta +\gamma )=-1$

hence, $c=(-1{)}^3+1.(-1{)}^2+2.(-1)+3$

$c=1$

Therefore the required equation is $x^3+2x^2+3x+1=0$