# When p (x) = x<su

We have,

x3 – ax2 + x

Let, p (x) = x3 – ax2 + x

And, x - a = 0

x = a

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

p (a) = (a)3 – a (a)2 + a

= a3 – a3 + a

= a

Hence, option B is correct

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