## Find sum of roots of polynomial

How would I go about approaching this problem?

Given the polynomial:
x^100 - 3x + 2 = 0

Find the sum 1 + x + x^2 + ... + x^99 for each possible value of x.
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 Quote by Cade How would I go about approaching this problem? Given the polynomial: x^100 - 3x + 2 = 0 Find the sum 1 + x + x^2 + ... + x^99 for each possible value of x.

If you meant that x is a root of the polynomial $X^{100}-3X+2$ , then
$$1+x+...+x^{99}=\frac{x^{100}-1}{x-1}=\frac{3x-3}{x-1}=3$$

DonAntonio
 Interesting, thanks, how did you derive that?

## Find sum of roots of polynomial

 Quote by Cade Interesting, thanks, how did you derive that?

First equality: sum of a geometric sequence.

Second equality: $x^{100}-3x+2=0\Longrightarrow x^{100}=3x-2$

Third equality: trivial algebra

DonAntonio
 Oh, I didn't realize the first part was the sum of a geometric series. Thanks for your help.
 Recognitions: Gold Member isn't a trivial solution to the equation equal to 1, then then sum would be greater than 3, This is the solution that makes the geometric sum equation impossible as you are dividing by zero.

 Quote by coolul007 isn't a trivial solution to the equation equal to 1, then then sum would be greater than 3, This is the solution that makes the geometric sum equation impossible as you are dividing by zero.

Indeed. So for $\,\,x=1\,\,,\,\,1+1^1+1^2+...+1^{99}=100\,\,$ , and for all the other roots it is what I wrote before.

Thanx.

DonAntonio