Descarte’s Rule determines how many real roots there are by counting the number of sign changes in the coefficients. I count three.

Rational zeroes indicates that x=±1 could be a root, so put x=1 into the polynomial:

1-2+1+1-2+1=0, so x-1 is a factor.

Use synthetic division to divide by the root:

1 | 1 -2 1 1 -2 1

__1 1 -1 0 1 -1__

1 -1 0 1 -1 | 0 = x⁴-x³+x-1.

Putting x=1 into this quartic also gives us 0, so x-1 is a factor, making (x-1)² a factor of the original polynomial.

Use synthetic division again:

1 | 1 -1 0 1 -1

__ 1 1 0 0 1__

1 0 0 1 | 0 = x³+1 = (x+1)(x²-x+1).

We can solve the quadratic: x=(1±√(1-4))/2=(1±i√3)/2.

The roots are therefore 1 (twice), -1, ½(1+i√3), ½(1-i√3). Two positive roots, one negative, and a pair of complex roots (conjugates).