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 ****.

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

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.


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.

  • $\begingroup$ understood, very well. $\endgroup$ – Bozo Vulicevic Jun 19 '15 at 18:27

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, $n=r+2s$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.