Polynomial Form for $f$ a Polynomial Such That $f(1)=0$

Question

What general form, as, for example $ax+by=c$, does the polynomial whose various forms are all evaluation at $1$ to be $0$?

$k_1\overbrace{(x-1)(x+a_1)(x+a_2)\cdots}^{\text{$n$ times}}+k_2\overbrace{(x-1)(x+b_1)(x+b_2)\cdots}^{\text{$n-1$ times}}+\cdots$

In other words, what is the form of the general polynomial $f$ such that $f(1)=0$? I think I need to know to show this.

If the answer to the question you ask here is not immediately obvious, then for the other question you have linked to, a more direct approach based on linear algebra should be easier and more straightforward. (of course, solving it by solving this question first is fine too... you should be able to solve it both ways!) — , Apr 17 '13 at 17:03

score 5 · Answer 1 · answered Apr 17 '13 at 16:27

5

$f(x)=g(x)(x-1)$, where $g$ is a polynomial. It is OK?

answered Apr 17 '13 at 16:27

Alex Ravsky

90,434

I guess I don't want an answer really... I'd just like some keen atypical insight. Perhaps I already know the "answer." – Trancot Apr 17 '13 at 17:25
2

@Trancot This is becoming complicated... Next time, please advertise your question with some mention like "Not wanting an answer really, keen atypical insights only", this will save us all some time. – Did Apr 18 '13 at 11:02

score 1 · Answer 2 · answered Apr 18 '13 at 10:59

Maybe you wanted this: a polynomial (function) $f(x)=a_nx^n+\cdots+a_1x_a+0$ satisfies $f(1)=0$ if an only if the coefficients satisfy $a_n+\cdots+a_1+a_0=0$, a single non-trivial linear equation. So the subspace of such polynomials (in any vector space of polynomials of degree${}<d$ for some $d>0$) has codimension$~1$ (and dimension $d-1$, one less than the dimension $d$ of the whole space).

Polynomial Form for $f$ a Polynomial Such That $f(1)=0$

2 Answers2