Difference between revisions of "1982 AHSME Problems/Problem 1"
(Created page with "By simply dividing <math>\frac{x^3-2}{x^2-2}</math> by using polynomial long division, you get : <math>x + \frac{2x-2}{x^2-2}</math>. Thus our answer will be <math>2x -2</mat...") |
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− | + | ==Problem== | |
− | Thus | + | |
+ | When the polynomial <math>x^3-2</math> is divided by the polynomial <math>x^2-2</math>, the remainder is | ||
+ | |||
+ | <math>\text{(A)} \ 2 \qquad | ||
+ | \text{(B)} \ -2 \qquad | ||
+ | \text{(C)} \ -2x-2 \qquad | ||
+ | \text{(D)} \ 2x+2 \qquad | ||
+ | \text{(E)} \ 2x-2</math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | Working out <math>\frac{x^3-2}{x^2-2}</math> using polynomial long division, we get <math>x + \frac{2x-2}{x^2-2}</math>. | ||
+ | Thus the answer is <math>2x -2</math>, for choice <math>\boxed{(E)}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME box|year=1982|before=First Question|num-a=2}} | ||
+ | |||
+ | {{MAA Notice}} |
Revision as of 23:01, 20 February 2019
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 .
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.