What should be subtracted from ${27 x}^{3} - 9 x^{2} - 6 x - 5$ to make it exactly divisible by $(3 x - 1)$ ?

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Solution

The correct option is B $- 7$
for exactly divisible by $(3 x - 1); x = \frac{1}{3}$ is
the root of $27 x^{3} - 9 x^{2} - 6 x - 5$
$\Rightarrow 1 - 1 - 6 (\frac{1}{3}) - 5 = - 7 ∴ (- 7)$ is to
be subtrated so that to be exactly divisible
by $(3 x - 1)$

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