Difference between revisions of "1951 AHSME Problems/Problem 18"
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== Problem == | == Problem == | ||
| − | The expression <math> 21x^2 | + | 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: |
<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> | ||
<math> \textbf{(D)}\ \text{some even number} \qquad\textbf{(E)}\ \text{zero}</math> | <math> \textbf{(D)}\ \text{some even number} \qquad\textbf{(E)}\ \text{zero}</math> | ||
| − | == Solution == | + | == 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> |
== See Also == | == See Also == | ||
Revision as of 20:21, 10 April 2013
Problem
The expression 
is to be factored into two linear prime binomial factors with integer coefficients. This can be one if 
is:
Solution
We can factor $21x^2 \plus{} ax \plus{} 21$ (Error compiling LaTeX. Unknown error_msg) as 
, which expands to 
. So the answer is 
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
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| All AHSME Problems and Solutions | ||
