# Find sum of roots of polynomial

1. May 1, 2012

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.

2. May 1, 2012

### DonAntonio

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

3. May 1, 2012

Interesting, thanks, how did you derive that?

4. May 1, 2012

### DonAntonio

First equality: sum of a geometric sequence.

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

Third equality: trivial algebra

DonAntonio

5. May 1, 2012

Oh, I didn't realize the first part was the sum of a geometric series. Thanks for your help.

6. May 2, 2012

### 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.

Last edited: May 2, 2012
7. May 2, 2012

### DonAntonio

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