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Difference between revisions of "1983 AHSME Problems/Problem 6"

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== Solution ==
 
== Solution ==
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<math>x^5(x+\frac{1}{x})\implies (x^6+\frac{1}{x^4})(1+\frac{2}{x}+\frac{3}{x^2})\implies x^6+J(x)</math>, so <math>6, \fbox{C}</math>
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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>.
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==See Also==
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{{AHSME box|year=1983|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 23:43, 19 February 2019

Problem 6

When $x^5, x+\frac{1}{x}$ and $1+\frac{2}{x} + \frac{3}{x^2}$ are multiplied, the product is a polynomial of degree.

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8$

Solution

We have $x^5\left(x+\frac{1}{x}\right)\left(1+\frac{2}{x}+\frac{3}{x^2}\right) = (x^6+x^4)\left(1+\frac{2}{x}+\frac{3}{x^2}\right) = x^6 + \text{lower order terms}$, where we know that the $x^6$ will not get cancelled out by e.g. a $-x^6$ term since all the terms inside the brackets are positive. Thus the degree is $6$, which is choice $\boxed{\textbf{(C)}}$.

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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