Question:
If the sum of the zeros of the polynomial $f(x)=2 x^{3}-3 k x^{2}+4 x-5$ is 6 , then the value of $k$ is
(a) 2
(b) 4
(c) $-2$
(d) $-4$
Solution:
Let $\alpha, \beta$ be the zeros of the polynomial $f(x)=2 x^{3}-3 k x^{2}+4 x-5$ and we are given that
$\alpha+\beta+\gamma=6$
Then,
$\alpha+\beta+\gamma=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$
$=-\frac{(-3 k)}{2}=\frac{3 k}{2}$
It is given that
$\alpha+\beta+\gamma=6$
Substituting $\alpha+\beta+\gamma=\frac{3 k}{2}$, we get
$\frac{+3 k}{2}=6$
$+3 k=6 \times 2$
$+3 k=12$
$k=\frac{12}{+3}$
$k=+4$
The value of $k$ is 4 .
Hence, the correct alternative is $(b)$