Difference between revisions of "1983 AHSME Problems/Problem 6"
Sevenoptimus (talk | contribs) (Further fixed the solution) |
Sevenoptimus (talk | contribs) (Improved clarity and added box at the bottom) |
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== Solution == | == Solution == | ||
− | We have <math>x^5\left(x+\frac{1}{x}\right)\left(1+\frac{2}{x}+\frac{3}{x^2}\right) = (x^6+x^4)\left(1+\frac{2}{x}+\frac{3}{x^2}\right) = 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>\ | + | We have <math>x^5\left(x+\frac{1}{x}\right)\left(1+\frac{2}{x}+\frac{3}{x^2}\right) = (x^6+x^4)\left(1+\frac{2}{x}+\frac{3}{x^2}\right) = 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> term since all the terms inside the brackets are positive. Thus the degree is <math>6</math>, which is choice <math>\boxed{\textbf{(C)}}</math>. |
+ | |||
+ | ==See Also== | ||
+ | {{AHSME box|year=1983|num-b=5|num-a=7}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 23:43, 19 February 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 term since all the terms inside the brackets are positive. Thus the degree is , which is choice .
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 |
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