Difference between revisions of "1951 AHSME Problems/Problem 18"
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== Solution == | == Solution == | ||
− | We can factor <math> 21x^2 + ax + 21</math> as <math>(7x+3)(3x+7)</math>, which expands to <math>21x^2+ | + | We can factor <math> 21x^2 + ax + 21</math> as <math>(7x+3)(3x+7)</math>, which expands to <math>21x^2+58x+21</math>. So the answer is <math> \textbf{(D)}\ \text{some even number}</math> |
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== Solution 2== | == Solution 2== | ||
Factoring <math>21x^2+ax+21</math> by grouping, we need to find some <math>b,c</math> such that <math>b\cdot c = 21^2</math>, and that <math>b+c=a</math>. | Factoring <math>21x^2+ax+21</math> by grouping, we need to find some <math>b,c</math> such that <math>b\cdot c = 21^2</math>, and that <math>b+c=a</math>. |
Revision as of 18:19, 5 June 2024
Contents
[hide]Problem
The expression is to be factored into two linear prime binomial factors with integer coefficients. This can be done if 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 | |
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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.