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

(Created page with "== Problem 6 == When <math>x^5, x+\frac{1}{x}</math> and <math>1+\frac{2}{x} + \frac{3}{x^2}</math> are multiplied, the product is a polynomial of degree. <math>\textbf{(A)}...")
 
(Cleaned up and added more explanation to the solution)
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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>
+
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>.

Revision as of 18:31, 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 $x^5(x+\frac{1}{x})(1+\frac{2}{x}+\frac{3}{x^2}) = (x^6+\frac{1}{x^4})(1+\frac{2}{x}+\frac{3}{x^2}) = x^6 + \text{lower order terms}$, 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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