Question:
If $\alpha, \beta, \gamma$ are the zeros of the polynomial $x^{3}-6 x^{2}-x+30$, then $(\alpha \beta+\beta \gamma+\gamma \alpha)=?$
(a) −1
(b) 1
(c) −5
(d) 30
Solution:
(a) $-1$
It is given that $\alpha, \beta$ and $\gamma$ are the zeroes of $x^{3}-6 x^{2}-x+30$.
$\therefore(\alpha \beta+\beta \gamma+\gamma \alpha)=\frac{\text { co-efficient of } x}{\text { co-efficient of } x^{3}}=\frac{-1}{1}=-1$
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