Factor $x^4-7x^2+1$ Is there a general method of factoring fourth order polynomials into a product of two irreducible quadratics? 
As I am reviewing on finding roots of polynomials in $\mathbb Z_n$ for abstract algebra, I am trying to factor the polynomial $x^4-7x^2+1$, and I was given the answer of $(x^2+3x+1)(x^2-3x+1)$. 
I was able to verify this of course by multiplying together the two irreducible quadratics, but I need to pretend that I never received the answer in the first place and ask for any hint in proceeding how to factor the fourth-order polynomial. Thanks.
It feels like I should use a method of "difference of squares" more than anything...

If you know the roots of your polynomial $p$, then you can factor it into quadratics by taking products of the linear terms. In the special case where $p$ looks like
$$
p(x) = ax^4 + bx^2 + c,
$$
you can substitute $u = x^2$ and then calculate the roots of
$$
au^2 + bu + c
$$
with the quadratic formula, assuming your ring is not characteristic $2$. With the roots of $u$, say they are $r,s$, you can write
$$
p = (u - r)(u-s).
$$
Your roots are going to be $\pm \sqrt r$, and $\pm \sqrt s$. Then $$
p = (x-\sqrt r))(x+\sqrt r)(x-\sqrt s )(x+\sqrt s).
$$
Do a little more multiplication to get into a factorization of quadratics.
