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Difference between revisions of "1986 AHSME Problems/Problem 24"

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==Solution==
 
==Solution==
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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
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<math>p(x)</math> must be a factor of <math>3(x^4+6x^2+25)-(3x^4+4x^2+28x+5)=14x^2-28x+70=14(x^2-2x+5)</math>.
  
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Therefore <math>p(x)=x^2 -2x+5</math> and <math>p(1)=4</math>.
  
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Answer: (D) 4
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The answer is <math>\fbox{(D) 4}</math>
  
 
== See also ==
 
== See also ==

Revision as of 16:18, 2 August 2016

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 (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
All AHSME Problems and Solutions

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