# Difference between revisions of "1960 AHSME Problems/Problem 31"

## Problem

For $x^2+2x+5$ to be a factor of $x^4+px^2+q$, the values of $p$ and $q$ must be, respectively:

$\textbf{(A)}\ -2, 5\qquad \textbf{(B)}\ 5, 25\qquad \textbf{(C)}\ 10, 20\qquad \textbf{(D)}\ 6, 25\qquad \textbf{(E)}\ 14, 25$

## Solution

Let the other quadratic be $x^2 + bx + c$, where $(x^2 + 2x + 5)(x^2 + bx + c) = x^4 + px^2 + q$. Multiply the two quadratics to get $$x^4 + (b+2)x^3 + (c + 2b + 5)x^2 + (2c + 5b)x + 5c$$ Since $x^4 + px^2 + q$ have no $x^3$ term and no $x$ term, the coefficients of these terms must be zero. $$b+2=0$$ $$b=-2$$ $$2c+5b=0$$ $$2c-10=0$$ $$c=5$$ Substitute $c$ and $b$ back to get $$x^4 + 6x^2 + 25$$ Thus, $p=6$ and $q=25$, so the answer is $\boxed{\textbf{(D)}}$.