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STEP II 1997 question 4 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1997 question 4 solution

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) p(1) = 3 \implies r(1) = 3. \quad p(2) = 1 \implies r(2) = 1. \quad p(3) = 5 \implies r(3) = 5. We know r(x) has degree < 3, so has the form Ax^2+Bx+C

r(1) = 3 \implies A+B+C = 3 (Eq 1)

r(2) = 1 \implies 4A+2B+C = 1 \implies 3A+B = -2 (Eq 2) (obtained by subtracting (Eq 1))

r(3) = 5 \implies 9A+3B+C = 5 \implies 5A+B = 4 (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

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