If $\alpha, \beta, y$ be the zeros of the polynomial $p(x)$ such that $(\alpha+\beta+\gamma)=3,(\alpha \beta+\beta y+y \alpha)=-$ 10 and $\alpha \beta y=-24$, then $p(x)=$ ?
(a) $x^{3}+3 x^{2}-10 x+24$
(b) $x^{3}+3 x^{2}+10 x-24$
(c) $x^{3}-3 x^{2}-10 x+24$
(d) None of these
(c) $x^{3}-3 x^{2}-10 x+24$
Given: $\alpha, \beta$ and $\gamma$ are the zeroes of polynomial $p(x)$.
Also, $(\alpha+\beta+\gamma)=3,(\alpha \beta+\beta \gamma+\gamma \alpha)=-10$ and $\alpha \beta \gamma=-24$
$\therefore p(x)=x^{3}-(\alpha+\beta+\gamma) x^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma$
$=x^{3}-3 x^{2}-10 x+24$
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