Factorization of polynomial $2x^4 - 2x^2 + 4$ I'm struggling with the polynomial factorization of the polynomial function $2x^4 - 2x^2 + 4$ in fields $\mathbb{R}, \mathbb{Q}, \mathbb{Z}, \mathbb{Z}_5, \mathbb{Z}_7$.
So far I have determined that it cannot be factored into something*function of degree 1 since it has no roots (used Newton calculus to find that it's "suspicious of extreme" points are positive as is an $f(0)=4>0$ making it a positive function).
That means it either is of irreducible factors already OR it is factored into multiplication of two polynomial functions of degree 2. 
I belove its the first case but don't know how to prove it since it ends in $5$ equations of $6$ variables (aka our function = $(ax^2 + bx + c)(ux^2 + vx + w)$). 
I'm not sure how to prove it/continue,
thanks for reply C:
 A: \begin{align}2(x^4-x^2+2)&=2\left[\left(x^2+\sqrt{2}\right)^2-(1+2\sqrt{2})x^2\right]\\&=2\left(x^2-\sqrt{1+\sqrt{2}}x+\sqrt{2\sqrt{2}}\right)\left(x^2+\sqrt{1+\sqrt{2}}x+\sqrt{2\sqrt{2}}\right)\end{align}
The scheme: $$u^2+a^2=(u+a)^2-2au$$
A: You can get disregard, for a while, the common factor $2$ and look for the factorization of $x^4-x^2+2$.
First let's look at $\mathbb{Q}$ and $\mathbb{R}$ that are related. Since the polynomial $y^2-y+2$ has no real roots, we can complete the square in a different way than usual:
$$
x^4-x^2+2=x^4+2\sqrt{2}x^2+2-(1+2\sqrt{2})x^2=
(x^2+\sqrt{2})^2-(\sqrt{1+2\sqrt{2}}x)^2
$$
from which you obtain the factorization over the reals and also that the polynomial is irreducible over $\mathbb{Q}$ (why?). However, the given polynomial $2x^4-2x^2+4$ is reducible over $\mathbb{Z}$ (why?).
Over the five element field you can observe that $y^2-y+2$ has no root by direct inspection, so $x^4-x^2+2$ has no root either. Hence its factorization can only be as the product of two monic degree $2$ polynomials:
$$
(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd
$$
so
$$
\begin{cases}
a+c=0\\
b+d+ac=-1\\
ad+bc=0\\
bd=2
\end{cases}
$$
The first equation yields $c=-a$ and the third becomes $a(d-b)=0$. If $a=0$, we obtain $b+d=-1$ and $bd=2$, so $b$ and $d$ should be roots of $y^2+y+2$ that has none. If $d=b$ we get $b^2=2$, which is impossible.
Can you work out the case of the seven element field? Note that $y^2-y+2=y^2+6y+9$ here.
