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...
 A: My first instinct in factoring is completing the square. $x^4-7x^2 + 1$ naturally becomes $(x^2-1)^2-5x^2$ or $(x^2+1)^2-9x^2$. Of the two the latter is a difference of squares and is thus more useful. Thus $$x^4-7x^2 + 1 = (x^2+1)^2-9x^2 = (x^2 + 1 + 3x)(x^2+1-3x)$$
A: Hint: $\;x^4-7x^2+1=x^2\left(x^2+\frac{1}{x^2}-7\right)=x^2\left(\left(x+\frac{1}{x}\right)^2-9\right)\,$ $=x^2\left(x+\frac{1}{x}+3\right)\left(x+\frac{1}{x}-3\right)$.
A: Hint:
Try with $$(x^2\pm1)^2-*x^2$$
The coefficient of $x^2$  has to be perfect square
A: 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.
A: Normally you have to solve a cubic to achieve this reduction (there are three different ways of pairing up the roots of the quartic to get two quadratic factors - and there are three solutions of the cubic). This does proceed (in one form at least) via the difference of two squares.
Here you can solve a quadratic in $x^2$ and then take square roots to get the four roots explicitly. Then you can choose how to pair them up. If you match any complex conjugate pairs you will get two real factors.
A: First we have to assume $x^2=z$, now our Equation is a quadratic equation , find the zeros (roots )of this equation. We will have two zeros in terms of $z$ , now put the value of $z$, so you can get the answer in $X$.those are the factors of the required equation .
A: There is a method. Trial division. For mnic quartic with no cube or linear term, and with square constant term,
$$ (x^2 - a x + b)(x^2 + ax + b).  $$
This time $b = \pm 1$
