$\displaystyle x^5 - 3x^4 + 2x^3 - 2x^2 + x + 1=0$

As I said, since you know that one of the roots is 1, use synthetic division:

Code:

1| 1 -3 2 -2 1 1
-- 1 -2 0 -2 -1
---------------------
1 -2 0 -2 -1 0

So your equation is the same as

$\displaystyle (x - 1)(x^4 - 2x^3 - 2x - 1) = 0$

Take the 2nd factor and solve for x:

$\displaystyle x^4 - 2x^3 - 2x - 1 = 0$

Subtract and add an x^2 on the left side:

$\displaystyle x^4 - 2x^3 - x^2 + x^2 - 2x - 1 = 0$

Factor by grouping **three** terms -- find the GCF for the first 3 terms, and do the same for the last 3 terms:

$\displaystyle x^2(x^2 - 2x - 1) + 1(x^2 - 2x - 1) = 0$

So this factors to

$\displaystyle (x^2 + 1)(x^2 - 2x - 1) = 0$

Set each factor to 0. For the first, there will be no real solutions. For the 2nd, use the quadratic formula.

EDIT: Ack! Too slow!