1983 AHSME Problems/Problem 18

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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{align*} f(y) &= x^4 + 5x^2 + 3 \\ &= (x^2)^2 + 5x^2 + 3 \\ &= (y - 1)^2 + 5(y - 1) + 3 \\ &= y^2 - 2y + 1 + 5y - 5 + 3 \\ &= y^2 + 3y - 1. \end{align*} Then substituting $x^2 - 1$, we get \begin{align*} f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\ &= x^4 - 2x^2 + 1 + 3x^2 - 3 - 1 \\ &= \boxed{x^4 + x^2 - 3}. \end{align*} The answer is (B).

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