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

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== Problem ==
 
== Problem ==
The expression <math> 21x^2 +ax +21</math> is to be factored into two linear prime binomial factors with integer coefficients. This can be one if <math> a</math> is:
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The expression <math> 21x^2 +ax +21</math> is to be factored into two linear prime binomial factors with integer coefficients. This can be done if <math> a</math> is:
  
 
<math> \textbf{(A)}\ \text{any odd number} \qquad\textbf{(B)}\ \text{some odd number} \qquad\textbf{(C)}\ \text{any even number}</math>
 
<math> \textbf{(A)}\ \text{any odd number} \qquad\textbf{(B)}\ \text{some odd number} \qquad\textbf{(C)}\ \text{any even number}</math>

Revision as of 19:00, 23 December 2015

Problem

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