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

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Solution

We divide $4 x^{4} - 2 x^{3} - 6 x^{2} + x - 5$ by $2 x^{2} + x - 1$ as shown in the below image:

From the division, we observe that the quotient is $2 x^{2} - 2 x - 1$ and the remainder is $- 6$ .

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

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