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

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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>\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 (ProblemsAnswer KeyResources)
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
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