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

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

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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 ?

Ray Vickson
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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

Ray Vickson
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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 ?

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

I do not see any question 2.

RGV

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