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Where g is a polynomial function of degree n-1

  1. Sep 16, 2012 #1
    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

    I don't know how to start. Please help

    Thank you in advance
     
    Last edited: Sep 16, 2012
  2. jcsd
  3. Sep 16, 2012 #2
    Re: Polynomials

    So, 0 is a root of this polynomial. What does that tell you? Factoring?
     
  4. Sep 16, 2012 #3
    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
     
  5. Sep 16, 2012 #4
    Re: Polynomials

    is that correct ?
     
  6. Sep 16, 2012 #5

    Ray Vickson

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    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
     
  7. Sep 16, 2012 #6
    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
     
  8. Sep 16, 2012 #7

    Ray Vickson

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    Re: Polynomials

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

    RGV
     
  9. Sep 16, 2012 #8
    Re: Polynomials

    ok I see. I do the same thing then for question 2 right ?
     
  10. Sep 16, 2012 #9

    Ray Vickson

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    Re: Polynomials

    I do not see any question 2.

    RGV
     
  11. Sep 16, 2012 #10
    Re: Polynomials

    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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