The remainder when $(6 x^{3} - 3 x + 2)$ is divided $(2 x - 3)$ is

81 / 4

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63 / 4

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71 / 4

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59 / 4

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Solution

The correct option is C $71 / 4$
When $(6 x^{3} - 3 x + 2)$ is divided by $(2 x - 3)$
Putting $2 x - 3 = 0$
We get,
$x = \frac{3}{2}$
Putting the value of $x$ is given expression
$= 6 \times {(\frac{3}{2})}^{3} - 3 \times \frac{3}{2} + 2 = 6 \times \frac{27}{8} - \frac{9}{2} + 2$
$= \frac{81}{4} - \frac{9}{2} + 2 = \frac{81 - 18 + 8}{4} = \frac{71}{4}$ .

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