# 1955 AHSME Problems/Problem 25

## Problem 25

One of the factors of $x^4+2x^2+9$ is: $\textbf{(A)}\ x^2+3\qquad\textbf{(B)}\ x+1\qquad\textbf{(C)}\ x^2-3\qquad\textbf{(D)}\ x^2-2x-3\qquad\textbf{(E)}\ \text{none of these}$

## Solution

We can test each of the answer choices by using polynomial division. $x^2 + 3$ leaves behind a remainder, and so does $x^2 - 3$.

In addition, $x + 1$ also fails the test, and that takes down $x^2 - 2x - 3$, which can be expressed as $(x + 1)(x - 3)$. That leaves $\boxed{(\textbf{E})}$

## Solution 2 (direct factorization)

Notice the leading and constant terms are begging us to create a binomial. So $$x^4 + 2x^2 + 9 = (x^4 + 6x^2 + 9) - 4x^2 = (x^2 + 3)^2 - (2x)^2 = (x^2 + 2x + 3)(x^2 - 2x + 3),$$ where both quadratics are irreducible (over the field of real numbers). Hence none of the given options is a factor. So the answer is $\boxed{(\textbf{E})}$

~VensL

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