Factorize in R[x] I have the polynomial $x^8+1$, I know that there's no root for solve this in $\Bbb R[x]$ but i want to factorize this to the minimal expression. This is possible or this is irreducible?
 A: Fun fact:
$$x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1).$$
This can be derived by setting $x^4+1$ equal to a product of two monic quadratics with unknown coefficients and then solving for said coefficients little by little. As a consequence,
$$x^8+1=(x^4+\sqrt{2}x^2+1)(x^4-\sqrt{2}x^2+1).$$
We can go further. For example, set
$$x^4+\sqrt{2}x^2+1=(x^2+ax+b)(x^2-ax+1/b),$$
yielding
$$\begin{cases}-a^2+b+1/b & =\sqrt{2} \\ a/b-ab & = 0 \end{cases} $$
Rule out $a=0$ to obtain $b=\pm1$ from the second equation, then plug those candidates into the first equation yielding $a=\sqrt{\pm2-\sqrt{2}}=\sqrt{2-\sqrt{2}}$ with $b=1$. The second factor of $x^8+1$ that is written above is similarly reducible. I leave the details as an exercise. The full factorization is
$$\begin{array}{lll} x^8+1 & = & \left(x^2+\sqrt{2-\sqrt{2}}x+1\right) \times \left(x^2-\sqrt{2-\sqrt{2}}x+1\right) \\ &  \times & \left(x^2+\sqrt{2+\sqrt{2}}x+1\right) \times\left(x^2-\sqrt{2+\sqrt{2}}x+1\right). \end{array}$$
over the real numbers $\bf R$. With the quadratic formula applied to the above you can get the roots to $x^8+1$ exactly (they are precisely the primitive $16$th roots of unity) in the form of nested radicals, and hence the full factorization in $\bf C$.
By the way, I should mention that the only nonlinear polynomials over $\bf R$ that are irreducible are quadratics with negative discriminant. This is because the nonreal roots of any real-coefficient polynomial can be paired off into conjugates, and then each conjugate pair of linear factors can be put together to obtain a real quadratic, thus every real-coefficient polynomial can be factored into real linear and quadratic factors.
Over $\bf Q$, as Belgi notes, a simple shift allows Eisenstein's criterion to apply.
A: The answer depends on $R$.
Example $1$: $R=\mathbb{C}$ and you know that any polynomial, including
the one in question, splits into linear factors 
Example $2$: $R=\mathbb{Q}$ denote $$f(x)=x^{8}+1$$ then $$f(x+1)=(x+1)^{8}+1=x^{8}+8x^{7}+28x^{6}+56x^{5}+70x^{4}+56x^{3}+28x^{2}+8x+2$$
is irreducible by Eisenstein with $p=2$ hence $f(x)$ is irreducible.
A: Over $\mathbb{C}$, this polynomial splits into linear factors of 8 of the 16 solutions to $x^{16} - 1 = 0$. The irreducible factors of this equation are the cyclotomic polynomials. 
http://en.wikipedia.org/wiki/Cyclotomic_polynomial
So $x^8 + 1$ is indeed irreducible over $\mathbb{Q}$, as it is a cyclotomic polynomial.
Over $\mathbb{R}$, this polynomial splits into 4 quadratic polynomials obtained by multiplying out conjugate pairs of 16th roots of unity which have order 16.
