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

$\textbf{(A)}\ x^4+5x^2+1\qquad \textbf{(B)}\ x^4+x^2-3\qquad \textbf{(C)}\ x^4-5x^2+1\qquad \textbf{(D)}\ x^4+x^2+3\qquad \textbf{(E)}\ \text{none of these}$

Solution

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

Solution 2

Let $y=x^2.$ We have that

$\begin{align*}f(x^2 + 1) &= x^4 + 5x^2 + 3 \\ &= y^2 + 5y + 3 \\ &= y^2 + 5y + 4 - 1 \\ &= (y+1)(y+4) -1 \\ &= (x^2 + 1)(x^2 + 4) - 1\end{align*}$

Thus, we have $f(n) = n(n+3)-1.$

If we plug in $n = x^2-1$ we have

$\begin{align*}f(x^2 - 1) &= (x^2 -1)(x^2 + 2)-1 \\ &= (x^4 + x^2 - 2) - 1 \\ &= \boxed{x^4 + x^2 -3} \end{align*}$

1983 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1983 AHSME Problems/Problem 18

Contents

Problem

Solution

Solution 2

See Also