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2020/21 AHSMC Part 1 Problem 14

Problem 14 of 16

The roots of a cubic polynomial p(x)p(x) are the squares of the three solutions of the equation x3−3x+1=0x^3 - 3x + 1 = 0. If p(0)=−4p(0) = -4, then p(4)p(4) is equal to

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Solution

Let q(x)=x3−3x+1=(x−a)(x−b)(x−c)q(x) = x^3 - 3x + 1 = (x-a)(x-b)(x-c). Then p(x)=k(x−a2)(x−b2)(x−c2)p(x) = k(x-a^2)(x-b^2)(x-c^2) and p(x2)=k(x2−a2)(x2−b2)(x2−c2)=k(x−a)(x−b)(x−c)(x+a)(x+b)(x+c)=−kq(x)q(−x). p(x^2) = k(x^2-a^2)(x^2-b^2)(x^2-c^2) = k(x-a)(x-b)(x-c)(x+a)(x+b)(x+c) = -kq(x)q(-x). Since p(0)=−4p(0) = -4 and q(0)=1q(0) = 1, one obtains k=4k = 4 and hence p(x2)=−4q(x)q(−x)p(x^2) = -4q(x)q(-x). Now p(4)=−4q(2)q(−2)=−4⋅3⋅(−1)=12. p(4) = -4q(2)q(-2) = -4 \cdot 3 \cdot (-1) = 12. The answer is (E).

University of Alberta