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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 ****.

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  • $\begingroup$ What is $t$? And what don't you understand? $\endgroup$ – anon Jun 19 '15 at 18:23
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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.

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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.

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  • $\begingroup$ understood, very well. $\endgroup$ – Bozo Vulicevic Jun 19 '15 at 18:27
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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, $n=r+2s$.

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