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The sum and product of an nth degree polynomial

  1. Oct 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose f(x) [tex]\in[/tex] Complex[x] is a monic polynomial of degree n with roots c1,c2,...cn. Prove that the sum of the roots is -a[tex]_{n-1}[/tex] and their product is (-1)[tex]^{n}[/tex]a[tex]_{0}[/tex]

    2. Relevant equations



    3. The attempt at a solution
    (x-c1)(x-c2)...(x-cn) = x[tex]^{n}[/tex] + (c1+c2+...+cn)x[tex]^{n-1}[/tex]....(c1*c2*....*cn)

    I just need a realistic proof this assumes too much
     
  2. jcsd
  3. Oct 14, 2008 #2

    Dick

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    In what way do you think that's assuming too much? Do you know the Fundamental Theorem of Algebra?
     
  4. Oct 14, 2008 #3
    but how do i know that (x-c1)(x-c2)...(x-cn) = xLaTeX Code: ^{n} + (c1+c2+...+cn)xLaTeX Code: ^{n-1} ....(c1*c2*....*cn)?
     
  5. Oct 14, 2008 #4

    Dick

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    Count powers of x. There's only one way to make x^n and x^0. There are n ways to make x^1. You just imagine multiplying it out.
     
  6. Oct 15, 2008 #5

    HallsofIvy

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    Because you know how to multiply polynomials?
     
  7. Oct 16, 2008 #6
    so thats a legit proof then?
     
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