Abstract Algebra, a question on polynomials . Theory : Every polynomial $P_n$ , of  n-th power over the field $R$- real numbers, can be written as a product of polynomials of the power $<3$, who's coefficients are real numbers.
Proof: $P_n(x)=a_0 + a_1(t)x+ ... a_nx^n(t)=a_n(x-x_1)(x-x_2)..(x-x_n)$
so: then it's written:**** $$(x-\alpha) (x+ \alpha)= x^2 +(\alpha+ \overline\alpha)x+ |\alpha|^2$$
also $\alpha+ \overline\alpha=2Re(\alpha)??$
so $$P_n=(x^2+p_1x+q_1)...(x^2+p_kx+q_k)...(x-x_{rm}), p_i, q_i \in R , i= \overline{1,k}$$ what do they mean by this? I get lose after ****.
 A: The proof starts with a complete factorization of the polynomial in the linear factors corresponding to its complex roots, real roots included.
It then recognizes that the non-real roots must appear in conjugate pairs, because the polynomial has real coefficients.
The product of the corresponding linear factors is a quadratic polynomials with real coefficients because these coefficients are twice the real part and the square of the absolute value of the roots.
Combining these results, you get a factorization of the original polynomial into real polynomials of degree 1 or 2.
A: If $\alpha$ is a non real root of $P$, then
$$\sum_{j=0}^na_j\alpha^j=0$$
But, since the coefficients $a_j$ are real numbers, taking conjugates, we get:
$$\sum_{j=0}^na_j\bar\alpha^j=0$$
that is, $\bar \alpha$ is a root of $P$, too.
Therefore,
$$P(x)=a_n(x-u_1)\cdots(x-u_r)(x-\alpha_1)(x-\bar\alpha_1)\cdots(x-\alpha_s)(x-\bar\alpha_s)$$
or
$$P(x)=a_n(x-u_1)\cdots(x-u_r)(x^2-2\Re\alpha_1+|\alpha_1|^2)\cdots(x^2-2\Re\alpha_s+|\alpha_s|^2)$$
Of course, $n=r+2s$.
A: In the complex numbers, conjugation (sending $a + bi \to a - bi$) is an automorphism, which means (among other things) that it is a ring homomorphism, that is:
$\overline{z + w} = \overline{z} + \overline{w}$, and:
$\overline{zw} = (\overline{z})(\overline{w})$.
Now suppose $p(x) \in \Bbb R[x]$, that is:
$p(x) = a_0 + a_1x +\cdots + a_nx^n$ with each $a_i \in \Bbb R$, and that $p(z) = 0$, for some $z = a + bi \in \Bbb C$.
Then $p(\overline{z}) = a_0 + a_1\overline{z} + \cdots + a_n\overline{z}^n$
$= a_0 + a_1\overline{z} + \cdots + a_n\overline{z^n}$ (since $(\overline{z})^k = \overline{z^k}$ for every natural number $k$),
$= \overline{a_0} + \overline{a_1}(\overline{z}) + \cdots + (\overline{a_n})\overline{z^n}$ (since for every real number $r$, we have $r = \overline{r}$)
$= \overline{a_0 + a_1z + \cdots + a_nz^n} = \overline{p(z)} = \overline{0} = 0$.
Hence, for any root $z$ of $p(x)$, we have $\overline{z}$ as a root, too (if $z$ is real, we only get one root perhaps).
So $(x - z)(x - \overline{z})$ is a factor of $p(x)$. But:
$(x - z)(x - \overline{z}) = x^2 - (z + \overline{z})x + z\overline{z}$.
Recalling that $z = a+bi$, we see:
$z\overline{z} = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 + b^2 \in \Bbb R$,
and $z + \overline{z} = a + bi + a - bi = 2a \in \Bbb R$. Hence $(x - z)(x - \overline{z}) \in \Bbb R[x]$, and we can factor it out of $p$.
So, real roots factor out as linear factors, and non-real roots factor out as quadratic factors.
