MHB Factoring Polynomial Equations

[Thetheorist]

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).

 

[kaliprasad]

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

 

[Thetheorist]

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.

 

[topsquark]

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

 

[Deveno]

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.