Difference between revisions of "1951 AHSME Problems/Problem 18"

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[[Category:Introductory Algebra Problems]]
 
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Revision as of 11:20, 5 July 2013

Problem

The expression $21x^2 +ax +21$ is to be factored into two linear prime binomial factors with integer coefficients. This can be one if $a$ is:

$\textbf{(A)}\ \text{any odd number} \qquad\textbf{(B)}\ \text{some odd number} \qquad\textbf{(C)}\ \text{any even number}$ $\textbf{(D)}\ \text{some even number} \qquad\textbf{(E)}\ \text{zero}$

Solution

We can factor $21x^2 \plus{} ax \plus{} 21$ (Error compiling LaTeX. Unknown error_msg) as $(7x+3)(3x+7)$, which expands to $21x^2+42x+21$. So the answer is $\textbf{(D)}\ \text{some even number}$

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AHSME Problems and Solutions

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