Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Find the answer

Let $f(x)=x^{3}+x+1$ . Suppose $g$ is a cubic polynomial such that $g(0)=-1$ , and the roots of $g$ are the squares of the roots of $f$ . Find $g(9)$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let $a, b, c$ be the zeros of $f$ . Then $f(x)=(x-a)(x-b)(x-c)$ . Then, the roots of $g$ are $a^{2}, b^{2}, c^{2}$ , so $g(x)=k(x-a^{2})(x-b^{2})(x-c^{2})$ for some constant $k$ . Since $a b c=-f(0)=-1$ , we have $k=k a^{2} b^{2} c^{2}=-g(0)=1$ . Thus, $g(x^{2})=(x^{2}-a^{2})(x^{2}-b^{2})(x^{2}-c^{2})=(x-a)(x-b)(x-c)(x+a)(x+b)(x+c)=-f(x) f(-x)$ . Setting $x=3$ gives $g(9)=-f(3) f(-3)=-(31)(-29)=899$ .

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