Question

If the sum of the zeroes of the polynomial .
$f\left(x\right)=2x^3-3k{x}^2+4x-5$ is $6$ then find the value of $k.$

Answer 1

$2x^3-3k{x}^2+4x-5=6$

If $\alpha ,\beta$ and $\gamma$ are the zeroes then,

$\alpha +\beta +\gamma =6$ [ Given ]

$\therefore$ $\frac{-b}{a}=6$ ---- ( 1 )

Since, $\alpha +\beta +\gamma =\frac{-b}{a}$

$\therefore$ $\frac{-b}{a}=\frac{-3k}{2}=\frac{coefficientof{x}^2}{coefficientof{x}^3}$

$\therefore$ $\frac{-b}{a}=6$ [ From ( 1 ) ]

$\therefore$ $\frac{3k}{2}=6$

$\therefore$ $3k=12$

$\therefore$ $k=4$