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Let $g \in \mathbb{C}[x]$ and $\omega \in \mathbb{C}$. Then:

  1. $\overline{g(\omega)} = \overline{g}(\overline{\omega})$

  2. If $g \in \mathbb{R}[x]$, then $\overline{g} = g$.

  3. Let $g \in \mathbb{R}[x]$ and $g(w) = 0$. Then $g(\overline{w}) = 0$.

(2) is quite obvious, since $z \in \mathbb{R} \Rightarrow z = \overline{z}$. However, I don't understand (1) and (3).

$\omega$ is a variable $\in \mathbb{C}$. So we plug some value in $\omega$ and then we find its conjugate and finally we calculate the value $g(\overline{w})$, right?

In general, I want to prove that in $\mathbb{R}[x]$ every polynomial can be expanded in product of polynomials, where degree of each polynomial $\leq 2$.

I know I have to use lemmas (1), (2), (3) and Fundamental theorem of algebra.

There are 3 cases to prove:

  1. $\deg f \leq 2$. Trivial. Proven.
  2. $\deg f > 2$ and $f$ has root $a \in R$. Then by Bezout's theorem $f = (x - a)q(x)$, where $\deg q = \deg f - 1$. Then I can apply induction on q, until it has no more real roots. Then I should use lemmas (1), (2), (3) and Fundamental theorem of algebra to say something about complex roots, but I stuck here.
  3. $\deg f > 2$ and $f$ has no real roots, and again repeat steps from (2) case.

Can you help me to prove the (1) and (3) lemmas and finish the theorem about polynomial expansion?

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$1.$ $\overline{a+b}=\overline{a}+\overline{b}$ and $\overline{ab}=\overline{a}\cdot\overline{b}$ for any two complex numbers $a$ and $b$. Therefore this possibility to split the complex conjugation exists for everything that can be composed from addition and multiplication of complex numbers, like the evaluation of a polynomial.

$3.$ follows from 1. and 2.: $$ 0=g(\omega) \Rightarrow 0=\overline 0 = \overline{g(\omega)}=\overline{g}(\overline\omega)=g(\overline\omega) $$

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  • $\begingroup$ So, if $f \in \mathbb{R}[x]$ it has root $\omega$, then $\overline{\omega}$ is also a root, then let $q(x) : = f(x) /( (x - \omega)(x - \overline{\omega}) )$ and I have to repeat this cycle for $q(x)$? $\endgroup$ – False Promise Feb 13 '17 at 17:26
  • $\begingroup$ Yes, that's how it works. You have to show that $(x-\omega)(x-\overline\omega)$ is in $\mathbb{R}[x]$, first. $\endgroup$ – Reinhard Meier Feb 13 '17 at 17:30
  • $\begingroup$ $(x-\omega)(x-\overline\omega) = x^2 - x(\omega + \overline{\omega}) + \omega \cdot \overline{\omega}$. $\omega \cdot \overline{\omega} =|\omega|^2 \in \mathbb{R}$. $\omega + \overline{\omega} \in \mathbb{R}$. And since we are in the field $\mathbb{R} \Rightarrow x \in \mathbb{R}$, right? $\endgroup$ – False Promise Feb 13 '17 at 17:41
  • $\begingroup$ Right. And this sentence exists only because stackexchange complains about short answers. $\endgroup$ – Reinhard Meier Feb 13 '17 at 17:43

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