Question

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

• A. 2
• B. 4
• C. -2
• D. -4

Answer 1

Answer: (B)

$4$

We know that the sum of the zeroes of the cubic polynomial $a{x}^3+b{x}^2+cx+d$ is $\frac{-b}{a}$

Given polynomial is $f\left(x\right)=2x^3-3k{x}^2+4x-5$
Here, $a=2,b=-3k,c=4$ and $d=-5$

Sum of zeroes $=\frac{-b}{a}$
$\Rightarrow$ Sum of zeroes $=\frac{-(-3k)}{2}$ [$\because a=2,b=-3k$]

$\Rightarrow 6=\frac{3k}{2}$ [$\because$ sum of zeroes $=6$]
$\Rightarrow 3k=6\times 2=12$
$\Rightarrow k=\frac{12}{3}$

$\Rightarrow$ $k=4$

Therefore, option B is correct.