Question

Find the value of $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }$ for the polynomial $x^3+2x-5$.

• A. $\frac{-1}{5}$
• B. $\frac{1}{5}$
• C. $\frac{-2}{5}$
• D. $\frac{2}{5}$

Answer 1

The correct option is D $\frac{2}{5}$

Given, $x^3+2x-5$

$\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{\beta \gamma +\gamma \alpha +\alpha \beta }{\alpha \beta \gamma }$

$a{x}^3+b{x}^2+cx+d\alpha \beta \gamma =\frac{-d}{a}\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}{x}^3+0x^2+2x-5a=1;b=0;c=2;d=-5\alpha \beta \gamma =\frac{-(-5)}{1}=5\alpha \beta +\beta \gamma +\gamma \alpha =\frac{2}{1}=2\therefore \frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{2}{5}$