I've shown that if $P(x) \in \mathbb{R}[X]$, then exist $Q_1(X), \dotsc, Q_k(X) \in \mathbb{R}[X]$ so that $P(X) = Q_1(X) \cdots Q_k(X)$ with $\deg Q_i \leq 2$ for all $1 \leq i \leq n$.

My proof - and every other I found - is based on the fundamental theorem of algebra. So I wonder, if it's possible to show this without use of the fundamental theorem? Explicitly, if it's possible to prove this within the real numbers? And if it's not - why?



It's pretty much equivalent to the fundamental theorem of algebra. It's easy to see that every quadratic polynomial with real coefficients splits in $\mathbf C$, so assuming the theorem you describe, we easily see that every real polynomial splits in $\mathbf C$. Conversely, as you already know, the fundamental theorem of algebra easily implies the other theorem.

  • $\begingroup$ Okay, that's clear to me - but is it possible to prove "my" theorem without using the fundamental theorem of algebra? $\endgroup$ – namsap Jan 25 '14 at 19:08
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    $\begingroup$ @gaussdorff Sure, but since the two theorems are equivalent, it will amount to proving the fundamental theorem of algebra itself. $\endgroup$ – Bruno Joyal Jan 25 '14 at 19:14
  • $\begingroup$ Okay, got it :) Thanks! $\endgroup$ – namsap Jan 25 '14 at 19:15
  • $\begingroup$ @gaussdorff My pleasure! $\endgroup$ – Bruno Joyal Jan 25 '14 at 19:41

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