Solving Polynomial Factors: Using Factor Theorem and Synthetic Division

In summary: Also, an easy way to find a root of a polynomial with rational coefficients is to plug in p/q for x, where p and q are relatively prime, and see if it equals 0. If so, then p/q is a root of the polynomial.In summary, the conversation discusses the use of the Factor Theorem and synthetic division to factor polynomials, specifically those with factors in the form (bx-a) instead of (x-a). It also raises the question of how to find factors with rational terms and the limitations of finding exact roots for polynomials of degree greater than 4. Possible methods for finding roots and factors are suggested, including plugging in rational values and factoring the leading coefficient and constant term.
  • #1
Hookflash
8
0
I was under the impression that any polynomial could be factored by using the Factor Theorem and synthetic division, but then I got to thinking: What if a polynomial has factors in the form (bx-a), as opposed to (x-a)? I can't find them with the Factor Theorem, so what do I do?

I imagine the answer will look something like this:

Given p(x) = (bx-a)(dx-c)(fx-e), I use the Factor Theorem to find the factors, (x-a/b), (x-c/d), (x-e/f), and bdf. Am I on the right track?
 
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  • #2
Hookflash said:
I was under the impression that any polynomial could be factored by using the Factor Theorem and synthetic division, but then I got to thinking: What if a polynomial has factors in the form (bx-a), as opposed to (x-a)? I can't find them with the Factor Theorem, so what do I do?

I imagine the answer will look something like this:

Given p(x) = (bx-a)(dx-c)(fx-e), I use the Factor Theorem to find the factors, (x-a/b), (x-c/d), (x-e/f), and bdf. Am I on the right track?
yes if it has a factor (bx-a) it also has a factor (x-a/b)
There are two problems
1)What field are you using
for instance if the field is real numbers
x^2+1 cannot be factored
The field of complex numbers is complete all polynomials can be completely factored
x^2+1=(x+i)(x-i)
2)How will you find the factors? For many polynomials finding and representing the factors can be a problem. For example with the 4 operations (+,-,*) and root extration you can find roots up to forth order equations (that is all of them some higher order equations can be factored easily, but not all) more functions are needed. Though it can be shown that the factors exist, it many be difficult to find them.
 
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  • #3
How will you find the factors?

I was wondering this myself, especially when I'm dealing with factors that have rational terms--i.e., (x-a/b). Those are a lot of terms to test via the Factor Theorem. I was thinking there must be a shortcut or something (aside from using my computer ;) ).
 
  • #4
Hookflash said:
I was wondering this myself, especially when I'm dealing with factors that have rational terms--i.e., (x-a/b). Those are a lot of terms to test via the Factor Theorem. I was thinking there must be a shortcut or something (aside from using my computer ;) ).

I'm wondering what "Factor Theorem" you are referring to. Unless the roots are rational, there will be an infinite number of terms to test.
Indeed, for polynomials of degree greater than 4, roots often cannot be written in terms of radicals! There is no way of finding, or even writing, exact roots.
 
  • #5
The factor theorem, at least according to Mathworld, is that if a is a root of the polynomial f, then (x - a) divides f.
 
  • #6
I just noticed that Hookflash did say that he was working with rational roots.

Hookflash, I don't know any way other than factoring the leading coefficient and constant term and constructing possible roots out of those.
 

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