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

Revision as of 19:19, 30 April 2015

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 $21x^2 + ax + 21$ as $(7x+3)(3x+7)$ , which expands to $21x^2+42x+21$ . So the answer is $\textbf{(D)}\ \text{some even number}$

1951 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == Solution ==
-We can factor <math> 21x^2 \plus{} ax \plus{} 21</math> as <math>(7x+3)(3x+7)</math>, which expands to <math>21x^2+42x+21</math>. So the answer is <math> \textbf{(D)}\ \text{some even number}</math>
+We can factor <math> 21x^2 + ax + 21</math> as <math>(7x+3)(3x+7)</math>, which expands to <math>21x^2+42x+21</math>. So the answer is <math> \textbf{(D)}\ \text{some even number}</math>
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Difference between revisions of "1951 AHSME Problems/Problem 18"

Revision as of 19:19, 30 April 2015

Problem

Solution

See Also