Question:
If $\alpha, \beta, \gamma$ are the zeros of the polynomial $2 x^{3}+x^{2}-13 x+6$, then $\alpha \beta y=?$
(a) −3
(b) 3
(c) $\frac{-1}{2}$
(d) $\frac{-13}{2}$
Solution:
(a) $-3$
Since $\alpha, \beta$ and $\gamma$ are the zeroes of $2 x^{3}+x^{2}-13 x+6$, we have:
$\alpha \beta \gamma=\frac{-(\text { constant term })}{\text { co-efficient of } x^{3}}=\frac{-6}{2}=-3$
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