My Math Forum Factoring a polynomial.

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 August 26th, 2012, 11:32 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 (x-a), 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, 02:14 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, 10:23 AM   #3
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Re: Factoring a polynomial.

Quote:
 Originally Posted by agentredlum 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
I understand the theorem( it's conclusion at least), I just don't know why it must be.

August 27th, 2012, 10:29 AM   #4
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Re: Factoring a polynomial.

Quote:
 Originally Posted by maxgeo I have always wondered why when factoring a polynomial say, 3x^2+1x^2+3
Why x^2 in 2nd term?

August 27th, 2012, 07:53 PM   #5
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Re: Factoring a polynomial.

Quote:
Originally Posted by maxgeo
Quote:
 Originally Posted by agentredlum 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
I understand the theorem( it's conclusion at least), I just don't know why it must be.
The key idea is divisibility, I attach excerpt from wiki link of 'an elementary proof' for convenience.

[attachment=0:2cr8jifg]Rational root theorem - Wikipedia, the free encyclopedia.png[/attachment:2cr8jifg]

Now let's work out an example using this proof.

Consider

$F(x)= x^2 - x - 6$

and suppose

$x= \frac{p}{q}$

is a rational root in lowest terms, that means p and q are co-prime, they have no common factors and makes $F(x)=0$ .

$F(\frac{p}{q})= \frac{p^2}{q^2} - \frac{p}{q} - 6 = 0$

move the constant -6 to the right hand side of the equation

$\frac{p^2}{q^2} - \frac{p}{q}= 6$

multiply by $q^2$

$p^2 - pq= 6q^2$

factor out $p$ on the left

$p(p-q)= 6q^2$

Now look at it very carefully... $p$ cannot divide $q^2$ because they are co-prime, so it must divide the coefficient of $q^2$ which is 6

The possible divisors of 6 are $\pm (1,2,3,6)$

now, using

$\frac{p^2}{q^2} - \frac{p}{q} - 6= 0$

move the leading term to the right hand side of the equation

$- \frac{p}{q} - 6= - \frac{p^2}{q^2}$

multiply by $-q^2$

$pq + 6q^2= p^2$

factor out $q$ from the left

$q(p + 6q)= p^2$

Now look at it carefully again. this time $q$ cannot divide $p^2$ because they are co-prime, so $q$ must divide the coefficient of $p^2$ which is 1

the possibble divisors of 1 are $\pm 1$

Putting it all together we see... IF there is a rational root $\frac{p}{q}$ in lowest terms, $p$ must divide the constant 6 (or -6, same divisors) and $q$ 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.

Attached Images
 Rational root theorem - Wikipedia, the free encyclopedia.png (94.5 KB, 167 views)

 August 28th, 2012, 01:01 AM #6 Member   Joined: Mar 2012 Posts: 60 Thanks: 0 Re: Factoring a polynomial. Yes, Thank you.
August 28th, 2012, 01:14 AM   #7
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Re: Factoring a polynomial.

Actually one more thing.

Quote:
 Originally Posted by agentredlum $p(p-q)= 6q^2$ Now look at it very carefully... $p$ cannot divide $q^2$ because they are co-prime, so it must divide the coefficient of $q^2$ which is 6.
This is assuming p and Q are integers so that (p-q) *p/p is also an integer? And since q is coprime with respect to p, Q/P cannot be an integer, therefore p must divide evenly into 6(the coefficient)?

August 28th, 2012, 02:12 AM   #8
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Re: Factoring a polynomial.

Quote:
Originally Posted by maxgeo
Actually one more thing.

Quote:
 Originally Posted by agentredlum $p(p-q)= 6q^2$ Now look at it very carefully... $p$ cannot divide $q^2$ because they are co-prime, so it must divide the coefficient of $q^2$ which is 6.
This is assuming p and Q are integers so that (p-q) *p/p is also an integer? And since q is coprime with respect to p, Q/P cannot be an integer, therefore p must divide evenly into 6(the coefficient)?
p is an integer, q is an integer, therefore (p-q) is an integer and (p-q) *p/p = (p-q) an integer

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 co-prime non-zero 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 co-prime.

not co-prime, violation

4*15 = 10*6

4,6 are not co-prime, and as a result 4 doesn't necessarilly divide 10, it doesn't...

things get tricky...

not co-prime, NON violation

4*6 = 12*2

4,2 are not co-prime, but 4 divides 12

However, if a,d are co-prime, then all the factors of a must be contained in c so a must divide c.

Hope this helps.

August 28th, 2012, 02:36 AM   #9
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Re: Factoring a polynomial.

Quote:
Originally Posted by agentredlum
Quote:
 Originally Posted by maxgeo Actually one more thing. hope this helps
Yes It does help, Thanks alto but I hope you don't mind if I ask one more slightly non-related question. Back when you posted the proof showing the Equation as-
a_(n)x^n+a_{n-1}x^{n-1}+\cdots+a_0 = 0\,\! What did the subscript represent. The a_n, a_(n-1), a_(n-2). It's not saying the coefficients have to be in descending order, is it?

August 28th, 2012, 02:58 AM   #10
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Re: Factoring a polynomial.

Quote:
Originally Posted by maxgeo
Quote:
Originally Posted by agentredlum
Quote:
 Originally Posted by maxgeo Actually one more thing. hope this helps
Yes It does help, Thanks alto but I hope you don't mind if I ask one more slightly non-related question. Back when you posted the proof showing the Equation as-
a_(n)x^n+a_{n-1}x^{n-1}+\cdots+a_0 = 0\,\! What did the subscript represent. The a_n, a_(n-1), a_(n-2). It's not saying the coefficients have to be in descending order, is it?
You are correct, the coefficients do not have to be in descending order. It is useful for identification purposes to put the POWERS of x in descending order so that the highest power of x, in this case n, is the leading term. Once you do it this way, it is clever to call the coefficient of $x^n$, $a_n$ because for polynomials with many terms we run out of letters of the alphabet.

continuing to match subscripts of coefficients to exponents of the variable, we get...

$F(x)= a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + ... + a_2x^2 + a_1x^1 + a_0x^0$

the last term, is just the constant $a_0$ because $x^0= 1$ for $x \ne 0$

hope this helps and you're welcome. Ask as much as you like, I don't mind at all.

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