# Talk:Polynomial Factor Theorem

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)