Factoring the quartic polynomial $x^4-2x^2+1$ Okay, I am practicing factoring for an upcoming assignment and I know that this is basic algebra, but I forgot how to attack this polynomial. Every method that I have used so far from simply guessing to using the quadratic formula to long division has failed me in replicating the answer. So either I am attacking this problem wrong up or there is another method that I just forgot.  
So here it is: $x^4-2x^2+1=0\\$
The 4 roots of the polynomial are $\pm 1,\pm 1$ (it only has 2 distinct roots).
My confusion is in how to get those values.
Any help in deriving the solution would be greatly appreciated.

$x^4-2x^2+1=0$ clearly has a root of $1$.  Since $\dfrac{x^4-2x^2+1}{x-1}= x^3+x^2-x-1$ you now have $$(x-1)(x^3+x^2-x-1)=0.$$
$x^3+x^2-x-1=0$ clearly has a root of $1$.  Since $\dfrac{x^3+x^2-x-1}{x-1}= x^2+2x+1$ you now have $$(x-1)(x-1)(x^2+2x+1)=0.$$
And so on.
