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April 16th, 2014, 08:53 PM  #1 
Newbie Joined: Apr 2014 From: Los Angeles Posts: 19 Thanks: 0  is this a polynomial equation?
It was my understanding that polynomial equations could not be divided by variables. But i came across this problem in rational expressions where supposedly one polynomial is divided by another. aren't a and b variables? imgur: the simple image sharer 
April 16th, 2014, 09:49 PM  #2 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra  
April 16th, 2014, 10:35 PM  #3 
Newbie Joined: Apr 2014 From: Los Angeles Posts: 19 Thanks: 0 
Well in this article mentions that polynomials can have variables, constans and exponents. but not fractional exponents or negative exponents and they cant' be divided by variables. I'm confused. :\ Polynomials 
April 16th, 2014, 10:59 PM  #4 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra 
I donâ€™t see what there is to be confused about. That page is simply saying that a polynomial cannot contain terms like $\dfrac2x$ (i.e. negative powers of the variable). However, if you are given a polynomial, thereâ€™s nothing to stop you from doing what you like with it. In particular you can divide a polynomial by another polynomial other than the zero polynomial to get whatâ€™s called a rational function. A rational function is an expression of the form $\displaystyle \frac{f(x)}{g(x)}$ where $f(x)$, $g(x)$ are polynomials with $g(x)\not\equiv0$. In abstract algebra the set of all rational functions with real coefficients is constructed as the field of fractions of the polynomial ring $\mathbb R[x]$. 
April 17th, 2014, 12:02 AM  #5 
Newbie Joined: Apr 2014 From: Los Angeles Posts: 19 Thanks: 0 
I just thought that if you had a variable as the denominator it wasn't a polynomial. When i saw this exercise i thought " ok one polynomial over another polynomial. The first one has an ab as denominator and the second has a 2a as denominator. a and b are both variables, therefore they can't be polynomials right?" Does it matter that the numerators are variables too? Btw how to you insert those neat functions like f(x)g(x)? Please teach me 
April 17th, 2014, 01:34 AM  #6  
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra  Quote:
Quote:
Last edited by Olinguito; April 17th, 2014 at 01:37 AM.  
April 17th, 2014, 04:24 PM  #7 
Newbie Joined: Apr 2014 From: Los Angeles Posts: 19 Thanks: 0 
Will do Thx Olinguito


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