1986 AHSME Problems/Problem 24

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Problem

Let $p(x) = x^{2} + bx + c$, where $b$ and $c$ are integers. If $p(x)$ is a factor of both $x^{4} + 6x^{2} + 25$ and $3x^{4} + 4x^{2} + 28x + 5$, what is $p(1)$?

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ 8$

Solution

$p(x)$ must be a factor of $3(x^4+6x^2+25)-(3x^4+4x^2+28x+5)=14x^2-28x+70=14(x^2-2x+5)$.

Therefore $p(x)=x^2 -2x+5$ and $p(1)=4$.

The answer is $\fbox{(D) 4}$.

See also

1986 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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