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April 16th, 2014, 08:53 PM   #1
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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?

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Attached Images poly equation.png (6.1 KB, 10 views) April 16th, 2014, 09:49 PM   #2
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Quote:
 Originally Posted by fsswim1 It was my understanding that polynomial equations could not be divided by variables.
Why not? 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 a-b 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
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Quote:
 Originally Posted by fsswim1 I just thought that if you had a variable as the denominator it wasn't a polynomial.
Thatâ€™s right. Rational functions are not polynomials (unless the denominator is a constant*). In other words:
1. $f(x)$ is a polynomial,
2. $g(x)$ is a polynomial, but
3. the whole thing $\displaystyle \frac{f(x)}{g(x)}$ is not a polynomial.
*Or more generally, unless the denominator is a factor of the numerator.

Quote:
 Originally Posted by fsswim1 Btw how to you insert those neat functions like f(x)g(x)? Please teach me If you browse the LaTeX Help section you might be able to find a few threads with instructions on how to insert $\LaTeX$.

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 Tags equation, polynomial Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Tzad Algebra 5 September 27th, 2013 09:15 AM Beiteltjie Advanced Statistics 1 October 5th, 2012 12:18 AM Lucida Algebra 8 May 23rd, 2012 07:11 PM stuart clark Algebra 3 March 3rd, 2011 06:02 AM jason1237 Algebra 3 October 30th, 2009 01:30 AM

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