1986 AHSME Problems/Problem 24

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 $\fbox{(D) 4}$ .

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1986 AHSME Problems/Problem 24

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