Proof that real coefficents polynomials of degree 3 and higher can be factorized into real linear and quadratic factors My book states that when you attempt to factorize a polynomial, one of three things may happen:
 - Being able to decompose the polynomial into linear factors using only real numbers.
 - Being able to decompose the polynomial into linear factors using only real numbers, but some of the factors may be repeated.
  - Being able to decompose the polynomial into linear and quadratic irreducible over the real numbers factors using only real numbers.
So, how can I proof that each real coefficents polynomial of degree 3 or higher can be factorized into linear and quadratic factors using only real numbers?
 A: A cubic equation with real coefficients has values as large (positive and negative) as you wish, so its graph must cross the real axis somewhere. That means it has at least one real root, so at least one real factor of the form $(x-r)$. The remaining factor will be quadratic. You can solve that with the quadratic formula and find two real roots or a pair of complex conjugate roots.
For polynomials of higher degree you use the fundamental theorem of algebra. You can look that up on wikipedia.
A: Yes, the fundamental theorem of algebra does indeed state that any complex-coefficient polynomial $P$ has $n$ roots in $\mathbb{C}$.
Now, a sketch of your proof would be:


*

*prove that if $\alpha$ is a root of $P$, $(X - \alpha)$ divides $P$ (hence: can write $P$ as the product $(X-\alpha_1)(X - \alpha_2)...(X - \alpha_n)$, where the $\alpha_i$ are its roots)

*prove that if $\alpha$ is a root, then so is $\bar{\alpha}$ (its complex conjugate), and then check that the quadratic $(X - \alpha)(X - \bar{\alpha})$ is a real polynomial


Those two points prove that your polynomial can be written as a product of real linear/quadratic factors :). (It's just a sketch, so if you have any question about how to prove each of those points, feel free to ask)
A: Consider the field automorphism ${\Bbb C}\rightarrow{\Bbb C}:z\mapsto \overline z$ mapping a complex number $z=a+ib$ to its complex conjugate $\overline z = a-ib$.
If $f(x)$ is a polynomial with real coefficients and $z$ is a complex root of it, then
complex conjugation yields $f(\overline z) = \overline{f(z)} = 0$ and so $\overline z$ is also a root of $f$. 
Thus $(x-z)(x-\overline z) = x^2 -(z+\overline z)x +z\overline z = x^2 - 2\Re(z)x + |z|^2$ is a factor of $f$ with real coefficients. From here the result follows.
