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August 26th, 2012, 10:32 PM  #1 
Member Joined: Mar 2012 Posts: 60 Thanks: 0  Factoring a polynomial.
I have always wondered why when factoring a polynomial say, 3x^2+1x^2+3 You find another polynomial such that when dividing (xa), a= (factors of the constant term)/(factor of leading coefficient.) so the possible factors are 3, 3/3, 3 and 3/3 Why is this?(that the only possible factors are Constant/leading coefficient) And is this only true when all coefficients and constants are whole numbers and not decimals. 
August 27th, 2012, 01:14 AM  #2 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Factoring a polynomial.
Hi maxgeo! It is called the 'rational roots theorem'. Read this wiki link, then let me know if you still have questions about it. http://en.m.wikipedia.org/wiki/Rational_root_theorem 
August 27th, 2012, 09:23 AM  #3  
Member Joined: Mar 2012 Posts: 60 Thanks: 0  Re: Factoring a polynomial. Quote:
 
August 27th, 2012, 09:29 AM  #4  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,136 Thanks: 1003  Re: Factoring a polynomial. Quote:
 
August 27th, 2012, 06:53 PM  #5  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Factoring a polynomial. Quote:
[attachment=0:2cr8jifg]Rational root theorem  Wikipedia, the free encyclopedia.png[/attachment:2cr8jifg] Now let's work out an example using this proof. Consider and suppose is a rational root in lowest terms, that means p and q are coprime, they have no common factors and makes . move the constant 6 to the right hand side of the equation multiply by factor out on the left Now look at it very carefully... cannot divide because they are coprime, so it must divide the coefficient of which is 6 The possible divisors of 6 are now, using move the leading term to the right hand side of the equation multiply by factor out from the left Now look at it carefully again. this time cannot divide because they are coprime, so must divide the coefficient of which is 1 the possibble divisors of 1 are Putting it all together we see... IF there is a rational root in lowest terms, must divide the constant 6 (or 6, same divisors) and must divide the coefficient of the leading term 1. This idea of divisibility can be extended to polynomials with higher degrees and more terms. Hope this helps.  
August 28th, 2012, 12:01 AM  #6 
Member Joined: Mar 2012 Posts: 60 Thanks: 0  Re: Factoring a polynomial.
Yes, Thank you.

August 28th, 2012, 12:14 AM  #7  
Member Joined: Mar 2012 Posts: 60 Thanks: 0  Re: Factoring a polynomial.
Actually one more thing. Quote:
 
August 28th, 2012, 01:12 AM  #8  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Factoring a polynomial. Quote:
for Q,P coprime, Q/P can be an integer if P = 1 or P = 1 Consider a simplified version of the divisibility idea... suppose a*b = c*d and a,d are coprime nonzero integers. then a must divide c because c must contain all factors of a for equality to hold. numerical examples: 4*6 = 8*3 4,3 are coprime so 4 must divide 8 and it does 5*14 = 35*2 5,2 are coprime so 5 must divide 35 and it does this idea doesn't work if a,d are not coprime. not coprime, violation 4*15 = 10*6 4,6 are not coprime, and as a result 4 doesn't necessarilly divide 10, it doesn't... things get tricky... not coprime, NON violation 4*6 = 12*2 4,2 are not coprime, but 4 divides 12 However, if a,d are coprime, then all the factors of a must be contained in c so a must divide c. Hope this helps.  
August 28th, 2012, 01:36 AM  #9  
Member Joined: Mar 2012 Posts: 60 Thanks: 0  Re: Factoring a polynomial. Quote:
a_(n)x^n+a_{n1}x^{n1}+\cdots+a_0 = 0\,\! What did the subscript represent. The a_n, a_(n1), a_(n2). It's not saying the coefficients have to be in descending order, is it?  
August 28th, 2012, 01:58 AM  #10  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Factoring a polynomial. Quote:
continuing to match subscripts of coefficients to exponents of the variable, we get... the last term, is just the constant because for hope this helps and you're welcome. Ask as much as you like, I don't mind at all.  

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