1983 AHSME Problems/Problem 18

Problem: Let $f$ be a polynomial function such that, for all real $x$ , $f(x^2 + 1) = x^4 + 5x^2 + 3.$ For all real $x$ , $f(x^2 - 1)$ is

Solution: (A) $x^4 + 5x^2 + 1$ (B) $x^4 + x^2 - 3$ (C) $x^4 - 5x^2 + 1$ (D) $x^4 + x^2 + 3$ (E) none of these Let $y = x^2 + 1$ . Then $x^2 = y - 1$ , so we can write the given equation as $\begin{aligned} f (y) & = x^{4} + 5 x^{2} + 3 \\ = (x^{2})^{2} + 5 x^{2} + 3 \\ = (y - 1)^{2} + 5 (y - 1) + 3 \\ = y^{2} - 2 y + 1 + 5 y - 5 + 3 \\ = y^{2} + 3 y - 1. \end{aligned}$ Then substituting $x^2 - 1$ , we get $\begin{aligned} f (x^{2} - 1) & = (x^{2} - 1)^{2} + 3 (x^{2} - 1) - 1 \\ = x^{4} - 2 x^{2} + 1 + 3 x^{2} - 3 - 1 \\ = x^{4} + x^{2} - 3 . \end{aligned}$ The answer is (B).

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1983 AHSME Problems/Problem 18