# When p (x) = x<su

We have,

x4 + 2x3 – 3x2 + x - 1

Let, p (x) = x4 + 2x3 – 3x2 + x - 1

And, x - 2 = 0

x = 2

It is given that, (x - 2) is a factor of p (x) so the remainder is equal to p (2)

p (2) = (2)4 + 2 (2)3 – 3 (2)2 + 2 - 1

= 16 + 16 – 12 + 2 – 1

= 34 – 13

= 21

Hence, option D is correct

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