Find the value of $k$ , if $2 x - 1$ is a factor of the polynomials. $6 x^{3} - x^{2} - 4 x + k$

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Solution

Let $P (x) = 6 x^{3} - x^{2} - 4 x + k$
$2 x - 1$ is a factor of $P (x)$
$\Rightarrow P (1 / 2) = 0$
$6 {(\frac{1}{2})}^{3} - {(\frac{1}{2})}^{2} - 4 (\frac{1}{2}) + k = 0$
$6 / 8 - 1 / 4 - 4 / 2 + k = 0$
$6 - 2 - 16 + 8 k = 0$
$8 k = 12$
$k = 3 / 2$

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