Need of an intuitive proof of an algebraic theorem While learning fraction decomposition I have introduced the following algebraic theorem but I don't know its proof.

Theorem:
  Any polynomial can be written as the product of linear factors and irreducible quadratic factors.

Can anyone please prove and explain the proof with intuition and examples?
 A: From the fundamental theorem of algebra, any polynomial $f$ with real coefficients can be factored as
$$f(x)=a(x-r_1)(x-r_2)\dots(x-r_n).$$
Since $f$ has real coefficients, $f(\bar x)=\overline{f(x)}$, so
$$a(\bar x-r_1)(\bar x-r_2)\dots(\bar x-r_n)=\bar a(\bar x-\bar r_1)(\bar x-\bar r_2)\dots(\bar x-\bar r_n).$$
Since $a$ is the leading coefficient, which is real by assumption, $\bar a=a$, and the roots on the left and right must be equal up to a permutation by the factor theorem (or unique factorization in $\Bbb C[x]$). Labeling those $r_i$ which are equal to their conjugates by $s_i$ and those which are not in pairs as $\alpha_i,\bar\alpha_i$, we get
\begin{align}
f(x)&=a(x-s_1)\dots(x-s_k)(x-\alpha_1)(x-\bar\alpha_1)\dots(x-\alpha_m)(x-\bar\alpha_m)\\
&=a(x-s_1)\dots(x-s_k)(x^2-2\Re[\alpha_1]x+|\alpha_1|^2)\dots(x^2-2\Re[\alpha_m]x+|\alpha_m|^2),\\
\end{align}
and this is our desired factorization, because the elements $x-s_i$ are linear real factors (since $s_i=\bar s_i$), and the quadratic terms $x^2-2\Re[\alpha_i]x+|\alpha_i|^2$ are real since $2\Re[\alpha_i],|\alpha_i|^2$ are real, and they are irreducible because their factorization $(x-\alpha_i)(x-\bar\alpha_i)$ contains complex coefficients.
For example, $x^5-1$ has the five roots $r_k=e^{2\pi ik/5}$, so its complex factorization is
$$x^5-1=(x-e^{-4\pi i/5})(x-e^{-2\pi i/5})(x-1)(x-e^{2\pi i/5})(x-e^{4\pi i/5}).$$
We pair up the conjugates $e^{2\pi i/5},e^{-2\pi i/5}$ and $e^{4\pi i/5},e^{-4\pi i/5}$, and the real root $1$ stays as it is, to get
\begin{align}
x^5-1&=(x-1)(x^2-(e^{2\pi i/5}+e^{-2\pi i/5})x+1)(x^2-(e^{4\pi i/5}+e^{-4\pi i/5})x+1)\\
&=(x-1)\left(x^2-2\cos\left(\frac{2\pi}5\right)x+1\right)\left(x^2-2\cos\left(\frac{4\pi}5\right)x+1\right),\\
\end{align}
which is our desired real factorization of $x^5-1$.
