How could I know that $X^4+1$ is $(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1)$? I thought that $X^4+1$ was irreducible, but in fact, $$X^4+1=(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1).$$
In general, how can I have the intuition of such a factorisation if I don't know it ?
 A: Hint
You can easily solve $X^4+1=0$ in $\mathbb C$ and identify which product of two monic are in $\mathbb R[X]$.
A: There's a sort of completion of the square that goes like this:
\begin{align}
x^4+1 & = \underbrace{(x^4+2x^2 + 1)}_\text{This is a square.} - \underbrace{(2x^2)}_\text{So is this.} \\[10pt]
& = \left( x^2+ 1 \right)^2 - (\sqrt 2\ x)^2 \\[10pt]
& = (x^2 + 1 - \sqrt 2\ x)(x^2 + 1  + \sqrt2\  x).
\end{align}
A: In fact $$(x^2 - x+1)(x^2 + x+1)$$ $$ = (x^2+1 -x)(x^2+1 +x)$$ $$= (x^2+1)^2 -x^2$$ (remember the  $\alpha^2-\beta^2 =(\alpha - \beta)(\alpha + \beta)$ formula? here $\alpha = x^2+1$ and $\beta = -x $) $$ = x^4 + x^2 +1$$
Well, if you really want to factor $x^4+1$, see here. Hope it helps. 
A: We know that $$a^4+4=a^4+4a^2+4-4a^2=(a^2+2)^2-(2a)^2=(a^2+2a+2)(a^2-2a+2)$$
This is an well known identity, most easily identifiable from the difference between two squares. It is called the Sophie Germain Identity= 
Putting in $x=\sqrt{2} a$, we have that $$x^2+1=(x^2+\sqrt{2} x+1)(x^2-\sqrt{2}x+1)$$
Though the first step is unnecessary, I added it as it is a generally useful formula. 
A: This is because the splitting field of $x^4+1$ is $K=Q(\zeta_8)$ and a quadratic subfield of that is $Q(\sqrt2)$.This may sound complicated if you are not versed in the terminology, but if you are interested,you can go through a Galois theory textbook, or more generally an abstract algebra textbook.
Basically, the quadratic subfields of K are $Q(\sqrt2),Q(\sqrt2i),Q(i), $ each corresponding to its Galois group (by the so called correspondence theorem).This theorem is what allows us to find these fields easily and what assures their finitude.Here are some factorisations of our polynomial: $x^4+1=(x^2+i)(x^2-i)=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1)=(x^2+\sqrt2ix-1)(x^2-\sqrt2ix-1).$
All three of these ways of factoring come from the aforementioned quadratic subfields of K.
EDIT: On an elementary note, $\zeta_8$, being the 8th root of unity, is actually equal to $\frac{i+1}{\sqrt2}$. Try to write the roots of 2,-2 and -1 as algebraic expressions of $\zeta_8$. This should directly give you that they are quadratic subfields of K.
A: If you're looking for real coefficients, every polynomial factors into a product of linear and quadratic terms.  Writing down the coefficients, on the other hand, can be nearly impossible.  Therefore, for your given polynomial, you know that it can be factored into quadratic terms.
There are deeper reasons behind the factorization in Galois theory over the rationals, but I'll go for the elementary approach - although we will need a detour through complex numbers.
To factor $x^4+1$, let's start by setting $y=x^2$.  Then, $x^4+1$ becomes $y^2+1$.  A quadratic can easily be factored, in this case we use the quadratic formula to get that the roots are $y=\pm i$.  Hence, this factors as
$$
y^2+1=(y-i)(y+i).
$$
Since $y=x^2$, we now know that
$$
x^4+1=(x^2-i)(x^2+i).
$$
Each of these are quadratics and can be factored with the quadratic formula.  In particular, $x^2-i$ has solutions $x=\pm\sqrt{i}=\pm\frac{\sqrt{2}}{2}\pm\frac{\sqrt{2}}{2}i$.  Therefore,
$$
x^2-i=\left(x-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)\left(x+\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right).
$$
Similarly, 
$$
x^2-i=\left(x-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)\left(x+\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right).
$$
Since we're looking for quadratics with real coefficients, we can try to pair these factors to get real coefficients after multiplying.  We can test out a few cases to find that the product
$$
\left(x-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)\left(x-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)=x^2-\sqrt{2}x+1.
$$
Similarly, the other pair results in $x^2+\sqrt{2}x+1$.
A: For me, the most intuitive way to see this is by factoring over the complex numbers.  Mainly notice that
$$x^4+1=0\implies x^4=-1\implies x=\operatorname{cis}(\pi(1+2k)/4),\quad k=0,1,2,3$$
where
$$\operatorname{cis}(\theta)=\cos(\theta)+i\sin(\theta)$$
Geometrically, it looks like this:

It is then easy to multiply these points back together to get
$$x^4+1=(x^2+\sqrt2x+1)(x^2+\sqrt2x-1)$$
A: There's a trick here, that is useful in other circumstances. I will  do it over the real numbers
$$
 x^{4} + 1 = x^{4} + 2 x^{2} + 1 - 2 x^{2} = (x^{2} + 1)^{2} - (\sqrt{2} x)^{2}
=
(x^{2} + 1 - \sqrt{2} x) (x^{2} + 1 + \sqrt{2} x).
$$
So it's just completing the square.
A: Iirc, Sophie Germaine in  Bending of Isotropic plates has these four roots generated (from the governing biharmonic equation in Theory of Plates  she had at first set up ) in  complex plane for $z^4+1= 0$ with the four roots $ (\pm \dfrac{1}{\sqrt2},  \pm \dfrac{i}{\sqrt2})$ given also here by Simply Beautiful Art.
A: we have a factoring rule: $a^4+b^4=(a^2-\sqrt2ab+b^2)(a^2+\sqrt2ab+b^2)$, then for question: $x^4+1=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1)$
A: There are many good answers already.  Here's a very elementary approach:
Let $f(x) = x^4 + 1$.
Over the real numbers, every polynomial factors into a product of linear and quadratic polynomials.  We know that $f(x) = 0$ has no solutions in the real numbers.  That means $f(x)$ doesn't have any linear factors.
So, it must have only quadratic factors.  Now since $f(x)$ is a polynomial of degree $4$, it must be the product of exactly $2$ quadratic polynomials:
$$
x^4 + 1 = (a_1 x^2 + b_1 x + c_1) (a_2 x^2 + b_2 x + c_2)
$$
From above we know that both factors are quadratic; so their $x^2$ terms are nonzero.  We might as well assume $a_1=1$ (because if not, we can divide by $a_1$ in the first factor, and multiply the second factor by $a_1$).  But then $a_2$ will be $1$ as well, because the $x^4$ term has coefficient $1$.
Similarly, $c_1c_2$ must equal $1$, so $c_2=1/c_1$ (and $c_1\neq 0$).
So let's rewrite the equation, using some new symbols for the coefficients:
$$
x^4 + 1  = (x^2 + ax + b)(x^2 + cx + 1/b)
$$
We do some rearranging:
$$
\begin{eqnarray}
x^4 + 1 & = &(x^2 + ax + b)(x^2 + cx + 1/b)\\
&=& x^4 + ax^3 + bx^2 + cx^3 +acx^2 + bcx + (1/b)x^2 + (a/b)x + 1\\
&=& x^4 + (a+c)(x^3) + (b + ac + 1/b)(x^2) + (bc + a/b)(x) + 1\\
\end{eqnarray}
$$
Since the coefficients on both sides have to be equal, we get three new equations:
$$
\begin{eqnarray}
0& =& a + c\\
0 &=& b + ac + 1/b\\
0& =& bc + a/b
\end{eqnarray}
$$
The first equation says $c = -a$, so let's substitute that into the other two equations:
$$
\begin{eqnarray}
0& =& b - a^2 + 1/b\\
0& = &-ab + a/b\\
\end{eqnarray}
$$
Or, multiplying both sides by $b$ in both equations:
$$
\begin{eqnarray}
a^2b &=& b^2 + 1\\
0& =& -ab^2 + a = a(-b^2 + 1) = a(1+b)(1-b)\\
\end{eqnarray}
$$
The first equation guarantees that $ a\neq 0$, so the second equation now says that $b=\pm 1$.  At this point the first equation says that $\pm a^2 = 2$, so in fact $a^2 = 2$ and $a = \pm \sqrt{2}$.  Now the first equation says $2b=2$, so $b=1$.
Case I: $a=\sqrt{2}$. Then we get
$$
x^4+1 = (x^2 + x\sqrt{2} + 1)(x^2 - x\sqrt{2} + 1)
$$
Case II: $a=-\sqrt{2}$.  Then we get
$$
x^4 + 1 = (x^2 - x\sqrt{2} + 1)(x^2 + x\sqrt{2} + 1)
$$
One case or the other has to be correct, but either way we get the same factorization (just written in a different order).  Therefore that factorization must be correct, and we're done.
