# Homework Help: Where g is a polynomial function of degree n-1

1. Sep 16, 2012

Let
Code (Text):
[ 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

Last edited: Sep 16, 2012
2. Sep 16, 2012

### Robert1986

Re: Polynomials

So, 0 is a root of this polynomial. What does that tell you? Factoring?

3. Sep 16, 2012

Re: Polynomials

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

4. Sep 16, 2012

Re: Polynomials

is that correct ?

5. Sep 16, 2012

### Ray Vickson

Re: Polynomials

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

6. Sep 16, 2012

Re: Polynomials

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

7. Sep 16, 2012

### Ray Vickson

Re: Polynomials

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

RGV

8. Sep 16, 2012

Re: Polynomials

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

9. Sep 16, 2012

### Ray Vickson

Re: Polynomials

I do not see any question 2.

RGV

10. Sep 16, 2012