Where g is a polynomial function of degree n-1

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Homework Help Overview

The discussion revolves around polynomial functions, specifically focusing on the implications of a polynomial having a root at zero and the conditions under which it can be expressed in a certain factored form. The original poster seeks assistance in demonstrating that if a polynomial function \( f(x) \) of degree \( n \) satisfies \( f(0) = 0 \), then it can be expressed as \( f(x) = xg(x) \), where \( g(x) \) is a polynomial of degree \( n-1 \).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of having zero as a root of the polynomial and the potential for factoring. There are attempts to clarify the relationship between the degrees of the polynomials involved and the conditions under which the polynomial can be expressed in a specific form. Questions arise about the clarity of the argument and the reasoning behind the factorization.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have provided insights into the implications of \( f(0) = 0 \) and the resulting factorization, while others express confusion about the clarity of the reasoning. There is no explicit consensus yet, but various lines of reasoning are being examined.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information they can use or the methods they can apply. The original poster has also introduced a second question regarding the conditions under which \( f(a) = 0 \) for some \( a \in \mathbb{R} \), which adds another layer to the discussion.

jawad1
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Let
Code:
 [ tex ]f(x)=\sum_{i=0}^n c_i x^i[ / tex ]
be an arbitrary polynomial function of degree n

Show that if f(0)=0 then either f is constant or f(x)=xg(x), where g is a polynomial function of degree n-1

I don't know how to start. Please help

Thank you in advance
 
Last edited:
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So, 0 is a root of this polynomial. What does that tell you? Factoring?
 


Since 0 is a root of f, the polynomial x divides the polynomial f. Therefore, there exists g such that xg(x)=f(x). However, if we write p the degree of g, degree of f= p+1 (since xg(x)=f(x)). Therefore if f equals zero, g equals zero. If not p=n-1
 


is that correct ?
 


jawad1 said:
is that correct ?

It seems you are making the argument much more obscure than it needs to be. Just put x = 0 in f(x) = c[0] + c[1]*x + c[2]*x^2 + ... + c[n]*x^n.

RGV
 


For the first part, I understood: we will get f(0)=c_0=0 which is a constant

However show that f(x)=xg(x) doesn't seem so obvious to me
 


jawad1 said:
For the first part, I understood: we will get f(0)=c_0=0 which is a constant

However show that f(x)=xg(x) doesn't seem so obvious to me

You don't see why
$$c_1x + c_2x^2 + \cdots + c_n x^n = x \cdot (\text{something})?$$

RGV
 


ok I see. I do the same thing then for question 2 right ?
 


jawad1 said:
ok I see. I do the same thing then for question 2 right ?

I do not see any question 2.

RGV
 
  • #10


Oh I am so sorry. I forgot to post it.

I need to show that if f(a)=0 for some a \in \mathbb{R}, the either f is constant or f(x)=(x-a)g(x), where g is a polynomial function of degree n-1
 

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