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All polynomials in $\Bbb{R}$ can be factored as a product of polynomials over $\Bbb R$ with degree one or two. This is a claim that I am having problems with. Can anyone prove it?

It's obvious if the degree is odd (remainder theorem and whatnot), but even polynomials are getting to me. I wanted to try using to fundamental theorem (as the problem came right after that section), but I don't know where to start and have little experience with complex numbers. We're just getting to Eisenstien's Criteria, to indicate how much/little I know.

Thanks.

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4 Answers

up vote 7 down vote accepted

Hint: for every complex root of a polynomial in $\mathbb R[x]$ the conjugate is a solution too of the same polynomial and every polynomial in $\mathbb C[x]$ factors as product of linear polynomial.

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Hint: $\mathbb{R}[x]$ is a UFD, and $\mathbb{C}$ is a degree two field extension of $\mathbb{R}$, which is algebraically closed. So what is the maximum degree that irreducible polynomials can have in $\mathbb{R}[x]$?

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Wow, quite simple, thank you. I see the proof before me. –  Pax Kivimae Jan 11 '13 at 18:42
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This is exactly the famous Fundamental Theorem of Algebra...

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If $\lambda$ is a real number, then it is obvious that its minimal polynomial is $x-\lambda$.

If $\lambda$ is complex-not-real, then $(x-\lambda)(x-\overline{\lambda})$ is a polynomial over the reals (check it!). So, the minimal polynomial over $\Bbb R$ of $\lambda$ has degree 2.

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