What must be subtracted from $4 x^{4} - 2 x^{3} - 6 x^{2} + x - 5,$ so that the result is exactly divisible by $2 x^{2} + x - 1$

- 5

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Solution

The correct option is B $- 6$
We have

$2 x^{2} + x - 1 = 0$

$\Rightarrow$ $2 x^{2} + 2 x - x - 1 = 0$

$\Rightarrow$ $2 x (x + 1) - 1 (x + 1) = 0$

$\Rightarrow$ $(2 x - 1) (x + 1) = 0$

$x = \frac{1}{2}, - 1$

Let $p$ is subtracted from

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

$\Rightarrow 4 x^{4} - 2 x^{3} - 6 x^{2} + x - 5 - p$

Therefore,

$\Rightarrow$ $4 (- 1)^{4} - 2 (- 1)^{3} - 6 (- 1)^{2} + (- 1) - 5 - p = 0$

$\Rightarrow$ $4 + 2 - 6 - 1 - 5 - p = 0$

$\Rightarrow$ $- p - 6 = 0$

$\Rightarrow$ $p = - 6$

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