Difference between revisions of "1982 AHSME Problems/Problem 1"
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Substituting an easy value, like <math>3</math>, you get <math>\frac{25}{7}</math>, with a remainder of 4. Then you just plug in x to get <math>E</math> | Substituting an easy value, like <math>3</math>, you get <math>\frac{25}{7}</math>, with a remainder of 4. Then you just plug in x to get <math>E</math> | ||
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==See Also== | ==See Also== |
Latest revision as of 12:27, 10 October 2022
Contents
Problem
When the polynomial is divided by the polynomial
, the remainder is
Solution
Working out using polynomial long division, we get
.
Thus the answer is
, for choice
.
Solution 2
Substituting an easy value, like , you get
, with a remainder of 4. Then you just plug in x to get
-dragoon
See Also
1982 AHSME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.