Difference between revisions of "1987 AHSME Problems/Problem 1"
(Created page with "==Problem== <math>(1+x^2)(1-x^3)</math> equals <math>\text{(A)}\ 1 - x^5\qquad \text{(B)}\ 1 - x^6\qquad \text{(C)}\ 1+ x^2 -x^3\qquad \\ \text{(D)}\ 1+x^2-x^3-x^5\qquad \tex...") |
Slackroadia (talk | contribs) (→Problem) |
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\text{(C)}\ 1+ x^2 -x^3\qquad \\ | \text{(C)}\ 1+ x^2 -x^3\qquad \\ | ||
\text{(D)}\ 1+x^2-x^3-x^5\qquad | \text{(D)}\ 1+x^2-x^3-x^5\qquad | ||
| − | \text{(E)}\ 1+x^2-x^3-x^6 </math> | + | \text{(E)}\ 1+x^2-x^3-x^6 </math> |
| + | ==Solution== | ||
| + | We multiply: <math>(1+x^2)(1-x^3) = 1 - x^3 + x^2 - x^5</math>. Thus the answer is <math>\boxed{D}</math> | ||
== See also == | == See also == | ||
Revision as of 18:33, 23 April 2017
Problem
equals
Solution
We multiply: 
. Thus the answer is 
See also
| 1987 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. 
