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The sum and product of an nth degree polynomial

by phyguy321
Tags: degree, polynomial, product
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phyguy321
#1
Oct14-08, 06:58 PM
P: 45
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
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Dick
#2
Oct14-08, 07:15 PM
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In what way do you think that's assuming too much? Do you know the Fundamental Theorem of Algebra?
phyguy321
#3
Oct14-08, 07:25 PM
P: 45
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)?

Dick
#4
Oct14-08, 10:23 PM
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The sum and product of an nth degree polynomial

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.
HallsofIvy
#5
Oct15-08, 05:18 AM
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Quote Quote by phyguy321 View Post
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)?
Because you know how to multiply polynomials?
phyguy321
#6
Oct16-08, 12:27 PM
P: 45
so thats a legit proof then?


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