Solving Fractions When Numerator Has Higher Degree

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Oblakastouf
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Alright... What are the rules for when the numerator (Top part) of the fraction has a higher degree than the lower part of the fraction.

Something + {*}

... As in what is that SOMETHING based on the numbers?

I have a midterm in like... 50 minutes and this is the only thing I need to know.
 
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Then you divide the two and you have an integer plus the remainder, where the numerator will have a smaller degree than the denominator. The process of dividing may take a little time but it is easy to check if you have the right answer (by simply multiplying your answer with one of the terms). Do you have a specific example?
 


VeeEight said:
Then you divide the two and you have an integer plus the remainder, where the numerator will have a smaller degree than the denominator. The process of dividing may take a little time but it is easy to check if you have the right answer (by simply multiplying your answer with one of the terms). Do you have a specific example?
Since you are talking about a fraction in which the numerator and denominator are polynomials, dividing will give you a polynomial (which corresponds to an integer in "regular" arithmetic) and remainder. The original fraction is equal to that polynomial plus a fraction with the remainder in the numerator, the original denominator as denominator.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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