1983 AHSME Problems/Problem 6: Difference between revisions
Katzrockso (talk | contribs) Created page with "== Problem 6 == When <math>x^5, x+\frac{1}{x}</math> and <math>1+\frac{2}{x} + \frac{3}{x^2}</math> are multiplied, the product is a polynomial of degree. <math>\textbf{(A)}..." |
Sevenoptimus (talk | contribs) Cleaned up and added more explanation to the solution |
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== Solution == | == Solution == | ||
<math>x^5(x+\frac{1}{x})\ | We have <math>x^5(x+\frac{1}{x})(1+\frac{2}{x}+\frac{3}{x^2}) = (x^6+\frac{1}{x^4})(1+\frac{2}{x}+\frac{3}{x^2}) = x^6 + \text{lower order terms}</math>, where we know that the <math>x^6</math> will not get cancelled out by e.g. a <math>-x^6</math> since all the terms inside the brackets are positive. Thus the degree is <math>6</math>, which is choice <math>\fbox{C}</math>. | ||
Revision as of 17:31, 26 January 2019
Problem 6
When 
and 
are multiplied, the product is a polynomial of degree.
Solution
We have 
, where we know that the 
will not get cancelled out by e.g. a 
since all the terms inside the brackets are positive. Thus the degree is 
, which is choice 
.
