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Let the remainder when p(x) is divided by (x-a) be r, so p(x) = (x-a)q(x)+r for some polynomial q. Then p(a) = (a - a)q(a) + r = r.

(i) . We know r(x) has degree < 3, so has the form

(Eq 1)

(Eq 2) (obtained by subtracting (Eq 1))

(Eq 3) (obtained by subtracting (Eq 2)).

Comparing (Eq 2) and (Eq 3), we see A = 3, so B = -11 and C = 11. So r(x) = 3x^2-11x+11.

(ii) If we didn't need a polynomial of degree (n+1), then p(x) = x would work, as p(a) = a for a = 0,...,n.

Analogously to (i), if we find a polynomial q of degree n+1 with q(a)=0 for a = 0,...,n; then q(x)+x is a polynomial of degree (n+1) with q(a) = a for a = 0,...,n.

Such a q(x) is x(x-1)(x-2)...(x-n). So our solution is x(x-1)(x-2)...(x-n)+x.

Solution by DFranklin