1960 AHSME Problems/Problem 31

Revision as of 18:44, 12 May 2018 by Rockmanex3 (talk | contribs) (Solution)

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 $\boxed{\textbf{(D)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 30
Followed by
Problem 32
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