Talk:Rational root theorem

One question

At the end of "A Proof" it says, "Thus... p/q is a root of f(x)." That's not so at all, is it? The proof starts with "Suppose p/q is a root..." Wouldn't it be best if the conclusion reminds the novice (me) of the conditional nature of the theorem: "IF a rational root exists, then..."?

Thanks.

--Dan kirshner (talk) 05:52, 23 August 2009 (UTC)[reply]

A few questions

I have a couple of questions and suggestions about this article. I'm new at this so please bear with me.

The parenthesized comment in the first line seems awkward to me because it suggests that the 'rational root test' is used to find the zeros, whereas the rational root theorem has another purpose. I suggest it be inserted after the parenthesis or removed altogether as superfluous.

As well, to be more specific, perhaps 'rational' should be inserted in 'constraint on solutions.'

The subscript on the constant coefficient is too large. It looks ugly and interferes with the reading.

In formal writing, I don't like a sentence to begin with an abbreviation. For example, further down the page, a sentence begins with 'E.g.'.

The exponents further along should be re-formatted to be consistent with the rest of the article.

'Rational Roots theorem' should be changed to 'rational root theorem' (although I guess the caps could stay on 'rational' since it is the name of something.

In the sentence 'Any polynomial ... must have a root in the set of real numbers' how about writing 'must have a real root' instead, and indicating the reason being that the roots come in conjugate pairs and reals are self-conjugate?

The formatting of subscripts and exponents in 'A Proof' needs improvement (I don't yet know how to do this).

I have a couple of general questions that I know belong in another place but don't know where.

On Wikipedia, are split infinitives okay? I don't like them and have a tendency to edit them out, but I don't want to appear pedantic or quarrelsome.

Do corrections of obviously misplaced modifiers require prior discussion (after all, such an edit usually does change the meaning of the sentence).

Thanks for any comments or help,

Aliotra (talk) 17:32, 17 March 2009 (UTC)[reply]

173.79.67.186 (talk) 00:20, 9 February 2009 (UTC)[reply]

In what sense are you dividing the set p by the set q? --RoseParks

That was an error in expression on my part. I was going to come back later on and fix it, but apparently someone else has done so already. --Apeiron

I removed the line

 (±1 cannot be a root because a₀/a_n is not integral.)

It so happens that neither 1 nor -1 is a root, but this is not implied by the rational root theorem. There is also no theorem saying that ±1 can only be a root if a₀/a_n is integral. For example, 3x³-x²-x-1 = 0 has a root x=1, even though -1/3 is not integral.

I also simplified the statement about the fundamental theorem of algebra. The versions "has a root" and "has n roots, if counted with multiplicities" are equivalent (i.e., each one easily implies the other -- from one root you can get to n roots with the above-mentioned Horner scheme), but I think the first version is easier to comprehend. Anyway, the second version can be found when following the link to the fundamental theorem of algebra.

Aleph4 09:41, 14 Apr 2004 (UTC)

I added the stipulation that a₀ should be nonzero, since if it is 0 then 0/1 is a rational root, but the numerator 0 is not a divisor of the constant term 0 - 0/0 is undefined.

The usual meaning of the phrase "m is a divisor of n" is that n=mk for some integer k. With this understanding, it is the case that 0 is a divisor of 0. In fact, every integer is a divisor of 0. The requirement that a₀ be nonzero can be dropped. However, the rational root theorem is not useful when the polynomial has zero constant term since there are infinitely many candidates to check. Because of this, it is worthwhile to have the comment that in this case 0 is a root.

Dadmo 20:15, 11 July 2008 (UTC)[reply]

We don't say what the one rational root is. Martin Packer (talk) 13:56, 25 September 2008 (UTC)[reply]

Merge

with Factor theorem?--Mongreilf (talk) 09:34, 10 April 2012 (UTC)[reply]

No, this is a different result that also happens to be about polynomials. Xnn (talk) 12:34, 27 November 2012 (UTC)[reply]