Difference between revisions of "1980 AHSME Problems/Problem 2"
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== Solution 1== | == Solution 1== | ||
| − | Expanding each factor yields <math>(x^{8}+...)(x^{9}+...)</math>. The degree of the first factor is <math>8</math>, while the degree of the second factor is <math>9</math>. Therefore, the degree of the polynomial is <math>8 + 9 = \boxed{ | + | Expanding each factor yields <math>(x^{8}+...)(x^{9}+...)</math>. The degree of the first factor is <math>8</math>, while the degree of the second factor is <math>9</math>. Therefore, the degree of the polynomial is <math>8 + 9 = \boxed{(\textbf{D})\ 17}</math> |
==Solution 2== | ==Solution 2== | ||
Latest revision as of 01:14, 14 July 2025
Contents
Problem
The degree of 
as a polynomial in 
is
Solution 1
Expanding each factor yields 
. The degree of the first factor is 
, while the degree of the second factor is 
. Therefore, the degree of the polynomial is 
Solution 2
Let 
be the degree of a polynomial in . Recall that for any polynomials 
and 
, and nonnegative integer 
, 
and 
.
So 
.
-mihirb, edited by j314andrews
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
