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 | + | The expression <math>21x^2+ax+21</math> is to be factored into prime binomial factors and without a numerical monomial factor. This can be done if the value ascribed to <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> | ||
Latest revision as of 12:29, 6 June 2024
Contents
Problem
The expression 
is to be factored into prime binomial factors and without a numerical monomial factor. This can be done if the value ascribed to 
is:
Solution
We can factor 
as 
, which expands to 
. So the answer is 
Solution 2
Factoring by grouping, we need to find some 
such that 
, and that 
.
Since 
, 
, and 
. So must be even. cannot be 
even number, since 
only has 4 odd factors, so the answer is
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
