Question

If $\alpha$, $\beta$, $\gamma$ are the zeroes of the polynomial $p\left(x\right)=6x^3+3x^2-5x+1$, find the value of $(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })$.

Answer 1

$p\left(x\right)=6x^3+3x^2-5x+1\alpha +\beta +\gamma =-\frac{b}{a}=\frac{-1}{2}\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}=\frac{-5}{6}\alpha \beta \gamma =-\frac{d}{a}=\frac{-1}{6}(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })=\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }=\frac{-5/6}{-1/6}=5$