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For every polynomial $p(x)$ of degree $n>3$, can we find a factorization of factors with degree at most $2$? (factorization in $\mathbb R$).

Example. $$x^5-1=(x-1)(x^4+x^3+x^2+x+1)=(x-1)(x^2+\frac{\sqrt{5}+1}{2}x+1)(x^2+\frac{-\sqrt{5}+1}{2}x+1)$$

In case $n$ even we need to talk:

If $n$ be odd and the coefficient of $x^n$ be positive (negative), we see $p(-\infty)=-\infty$ and $p(+\infty)=+\infty$ $\Big(p(-\infty)=+\infty$ and $p(+\infty)=-\infty\Big)$ then it has a root and we reduce degree to case $n-1$.

Peter
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Nosrati
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    Yes, the reason is that every polynomial with odd degree has a real root and if $z=a+bi$ is a root, then $z=a-bi$ is also a root. – Peter Jan 12 '16 at 08:42
  • Also true for $n=3$. –  Jan 12 '16 at 08:57
  • If you mean ‘Is there an algorithm?’ to factor any polynomial’, the answer is no, by Galois theory, if $n>4$. If you mean ‘ Does there exist a factorisation?’, the answer is yes by D'Alembert-Gauß theorem. – Bernard Jan 12 '16 at 09:48
  • @Bernard If there is not an algorithm to factor any polynomials, I need tell me more or give me a reference about Galois theory. – Nosrati Jan 12 '16 at 10:29
  • One of the best references, in my opinion, is David A. Cox, Galois Theory (Wiley): it has historical notes and a computational point of view. See Theorem 8.5.6 about non-solvability by radicals of the general equation of degree $n>5$. At the end of this chapter he mentions a proof of this fact without Galois theory by R. G. Ayoub, in the American Mathematical Monthly. – Bernard Jan 12 '16 at 10:48

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Yes, this follows from (and is essentially equivalent to) the fundamental theorem of algebra. By the fundamental theorem of algebra, $p(x)$ has a complex root $a+bi$ for some $a,b\in\mathbb{R}$. If $b=0$, this is a real root and we're done, so suppose $b\neq 0$. Since $p(x)$ has real coefficients, $\overline{p(x)}=p(\bar{x})$ for any $x\in\mathbb{C}$, where the bar denotes complex conjugation. It follows that $0=\overline{p(a+bi)}=p(a-bi)$, so $a-bi$ is also a root of $p$. Now note that $$q(x)=(x-(a+bi))(x-(a-bi))=x^2-2ax+(a^2+b^2)$$ has real coefficients, and divides $p(x)$ since $a+bi$ and $a-bi$ are both roots of $p(x)$. Thus $p(x)$ has a quadratic factor with real coefficients, and by induction on degree we're done.

Eric Wofsey
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