Question:
If two zeroes of the polynomial $x^{3}+x^{2}-9 x-9$ are 3 and $-3$, then its third zero is
(a) $-1$
(b) 1
(c) $-9$
(d) 9
Solution:
Let $\alpha=3$ and $\beta=-3$ be the given zeros and $\gamma$ be the third zero of the polynomial $x^{3}+x^{2}-9 x-9$ then
By using $\alpha+\beta+\gamma=\frac{-\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$
$\alpha+\beta+\gamma=\frac{-1}{1}$
$\alpha+\beta+\gamma=-1$
Substituting $\alpha=3$ and $\beta=-3$ in $\alpha+\beta+\gamma=-1$, we get
$3-3+\gamma=-1$
$\gamma=-1$
Hence, the correct choice is $(a)$
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