Question:
If the product of two zeros of the polynomial $f(x)=2 x^{3}+6 x^{2}-4 x+9$ is 3 , then its third zero is
(a) $\frac{3}{2}$
(b) $-\frac{3}{2}$
(C) $\frac{9}{2}$
(d) $-\frac{9}{2}$
Solution:
Let $\alpha, \beta, \gamma$ be the zeros of polynomial $f(x)=2 x^{3}+6 x^{2}-4 x+9$ such that $\alpha \beta=3$
We have,
$\alpha \beta \gamma=\frac{\text { Constant term }}{\text { Coefficient of } x^{3}}$
$=\frac{-9}{2}$
Putting $\alpha \beta=3$ in $\alpha \beta \gamma=\frac{-9}{2}$, we get
$\alpha \beta \gamma=\frac{-9}{2}$
$3 \gamma=\frac{-9}{2}$
$y=\frac{-9}{2} \times \frac{1}{3}$
$\gamma=\frac{-3}{2}$
Therefore, the value of third zero is $\frac{-3}{2}$.
Hence, the correct alternative is $(b)$.