1983 AHSME Problems/Problem 6

Revision as of 17:48, 26 January 2019 by Sevenoptimus (talk | contribs) (Further fixed the solution)

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.

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


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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