Talk:Polynomial Factor Theorem

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Can the original version of the theorem as applied to a polynomial of $\R$ be added a corollary, please? It's much less intimidating for those amateurs like me whose sphere of knowledge is limited. --GFauxPas 08:45, 2 December 2011 (CST)

Yep. --prime mover 13:14, 2 December 2011 (CST)
Thank you! --GFauxPas 13:22, 2 December 2011 (CST)

In the Corollary statement, shouldn't it be written "$P(\xi_1)=P(\xi_2)=\ldots=P(\xi_n)=0$" instead of "$P(\xi_1)=P(\xi_2)=\ldots=P(\xi_n)$" ? Same thing in the Proof section when you write "We can then apply this result to the situation where $P(\xi_1)=P(\xi_2)=\ldots=P(\xi_n)$"... shouldn't it be written "We can then apply this result to the situation where $P(\xi_1)=P(\xi_2)=\ldots=P(\xi_n)=0$" ?

I guess.

Also why do you switch to capital letters for $x$ in the third sentence of the Proof section ?

cos I'm a dick.

Also, although it is not really an inconsistency, you introduce the polynomial by $P(x)$, then in the first part of the Proof section you drop the $x$ dependence of the polynomials $P$, $Q$ and $R$ and then you re-introduce the dependencies in the last part of the proof. Also shouldn't you point out that $Q$ and $R$ are in $K[x]$ ? Finally I think that it is possible to be more precise about the constant $k \in K$ : isn't it the leading coefficient of $P(x)$ ? That is, if

$P(x)=a_0+a_1 x+a_2 x^2+\cdots+a_n x^n=k(x-\xi_1)(x-\xi_2) \cdots (x-\xi_n)=k x^n + \cdots + (-1)^n k \xi_1 \xi_2 \cdots \xi_n$,

then can't we use the linear independence of the functions $1, x, x^2, \ldots, x^n$ (the Wronskian for these functions is different from 0) to conclude that coefficients must match and that in particular $k=a_n$ ? Precisely, if $1, x, x^2, \ldots, x^n$ are linearly independent, then $(a_0-(-1)^n k \xi_1 \xi_2 \cdots \xi_n)+\cdots+(a_n-k)x^n=0 \implies a_n-k=0 \implies a_n=k$ ? --Frades

dunno
Looks like it got a rewrite last year and from that point on it has failed to make sense. --prime mover 18:43, 1 June 2012 (EDT)
If you want to edit the proof to make it hold together better or put up an alternative proof, feel free! Just don't delete the existing content without discussing it:) --Alec (talk) 22:07, 1 June 2012 (EDT)