What should be added to $(x^{4} - 1)$ so that it is exactly divisible by $(x^{2} + 2 x + 1)$ ?

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Solution

It is given that the dividend is $p (x) = x^{4} - 1$ , the divisor is $g (x) = x^{2} + 2 x + 1$ , therefore, the division is as shown above.

Therefore, from the division, we get that the remainder is $r (x) = (- 4 x - 4)$ that is $- r (x) = 4 x + 4$ .

Hence, $4 x + 4$ should be added to $x^{4} - 1$ so that the resulting polynomial is exactly divisible by $x^{2} + 2 x + 1$ .

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What should be added to (x4−1) so that it is exactly divisible by (x2+2x+1)?

What should be added to $(x^{4} - 1)$ so that it is exactly divisible by $(x^{2} + 2 x + 1)$ ?