Factoring Polynomial Equations

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    Factoring Polynomial
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Discussion Overview

The discussion centers on the possibility of determining whether a polynomial has rational zeroes or is irreducible over the rationals without resorting to trial and error. Participants explore methods for assessing polynomials, particularly in the context of time constraints during tests.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant inquires whether there is a method to determine if a polynomial is unfactorable using whole numbers without trial and error.
  • Another participant suggests that since the coefficients of the polynomial are even, there may not be an integer solution.
  • A participant asserts that there is no general method to determine if a polynomial with integer coefficients is irreducible over the rationals, but mentions some potential shortcuts.
  • Shortcuts proposed include reduction modulo a prime, Eisenstein's criteria, and using calculus to analyze the polynomial's behavior.
  • One participant emphasizes the complexity of factoring polynomials, particularly those of degree four and higher, noting that there is no general solution for degree five or higher.
  • A later reply highlights a specific case where an integer solution leads to a contradiction, reinforcing the idea of irreducibility.

Areas of Agreement / Disagreement

Participants generally agree that there is no straightforward method to determine irreducibility for all polynomials, and multiple competing views on potential shortcuts and methods remain. The discussion does not reach a consensus on a definitive approach.

Contextual Notes

Limitations include the dependence on specific criteria for irreducibility and the complexity of polynomials of higher degrees. The discussion does not resolve the effectiveness of the proposed shortcuts.

Thetheorist
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I would like to know if it is possible to determine if a polynomial has rational zeroes, or, in other words, is unfactorable using whole numbers.

For example 4x^3+2x^2-4x+25.

I know you can use trial and error to sub in the factors of 25, and I understand the rational root theorem. However, I was wondering if there is a way to look at that equation and determine right away that it is not factorable using whole numbers without going through the process of trial and error subbing (helps to save time on tests).
 
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For us to provide help we expect you to mention what you have tried. Secondly because the coefficient of x^3 is 4 and not one you should try $\pm 5$,$\pm 1$,$\pm \frac{5}{2}$, $\pm \frac{5}{4}$,$\pm \frac{1}{2}$,$\pm \frac{1}{4}$. (that is factors of 5 divided by factors of 4)
by quick inspection as coefficient of all terms except constant are even so there is no integer solution
 
Sorry if I was unclear, but my question is not about finding the roots for that specific function. My question is, can you look at a polynomial in standard form and determine right away that it will have rational roots? As of right now, I go through trial and error of subbing in whole numbers, and if that doesn't work then I assume it has rational roots. But on a test, there is limited time to do this.
 
Thetheorist said:
Sorry if I was unclear, but my question is not about finding the roots for that specific function. My question is, can you look at a polynomial in standard form and determine right away that it will have rational roots? As of right now, I go through trial and error of subbing in whole numbers, and if that doesn't work then I assume it has rational roots. But on a test, there is limited time to do this.
As a general theorem you have the one word answer: Nope.

-Dan
 
What you are asking is-is there a general way to determine if a polynomial with integer coefficients is *irreducible* over the rationals.

No, but there are some "shortcuts" that sometimes work:

1. Reduction modulo $p$, a prime. If it factors over the integers, it will still factor over the integers mod $p$, and sometimes this is easier to determine.

2. Eisenstein's criteria: If a prime $p$ divides every coefficient but the leading one, and $p^2$ does not divide the constant term, the polynomial is irreducible (and thus has no factorizations at all, much less any roots).

3. One can use calculus to determine where the "humps" and "valleys" lie, and thus determine which intervals are "all one sign". This isn't always the easiest approach, but it can sometimes narrow down the search enough to save time on the rational root test.

In general, determining whether a polynomial factors over a field is a "hard" question. Polynomials of degree $4$ are already extremely complicated to factor, and for polynomials of degree $5$ of higher, there is no "general" solution.

kaliprasad's comment is worth a few extra words, in this case we have:

$2x(2 - x - 2x^2) = 25$

if $x$ was an integer, we would have $25$ as an even number, contradiction.
 

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