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

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== Solution ==
 
== Solution ==
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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>.
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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> 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 18:48, 26 January 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 $-x^6$ since all the terms inside the brackets are positive. Thus the degree is $6$, which is choice $\fbox{C}$.

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