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Takumi Murayama says "Every polynomial in $\mathbb R[x]$ of degree at least 3 has a real root, and therefore is not irreducible". I think I understand why it is not irreducible, but what's the real root of $f(x)=(x^2+1)(x^2+2)$?

If he is right, then why?

If it is wrong, then what is probably meant? I think this has something to do with complex roots in pairs.

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    I believe the writer intended to add that the degree was odd. Otherwise, the claim is false for the reason you report. There are polynomials of arbitrarily high even degree with no root, $p_n(x)=(x^2+1)^n$ has degree $2n$ and has no real root, for example. – lulu Dec 22 '18 at 12:06
  • Yes, lulu is right; see this post. – Dietrich Burde Dec 22 '18 at 12:08
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    The link from Murayama you attach is extremely long. On what page does that quote appear? – lulu Dec 22 '18 at 12:09
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    The author's claim (on p. 39) is wrong. What he meant is that every polynomial of degree three or more has irreducible factors which are at most quadratic: "So, every maximal ideal is of the form (f) for f a linear polynomial or an irreducible quadratic polynomial." – Michael Hoppe Dec 22 '18 at 12:09
  • @MichaelHoppe I thought something like that. Why don't you post as an answer? Thank you! –  Dec 25 '18 at 20:30
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    @JackBauer Well, in my opinion a comment is sufficient. – Michael Hoppe Dec 25 '18 at 20:44
  • @lulu p. 39 last paragraph –  Dec 25 '18 at 20:48
  • @TakumiMurayama Then I'll just cite the solution as anonymous source? Thank you for your solutions! –  Dec 30 '18 at 05:02

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From the comments

The author's claim (on p. 39) is wrong. What he meant is that every polynomial of degree three or more has irreducible factors which are at most quadratic: "So, every maximal ideal is of the form (f) for f a linear polynomial or an irreducible quadratic polynomial." – Michael Hoppe Dec 22 at 12:09