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After encountering many numerical problems related to finding real roots of a polynomial, I have fixed a simple path:

  1. For an odd-degree polynomial: complex roots come in pairs. If a polynomial is of degree $d$ which is odd then the polynomial must have at least one real root.

  2. For a even-degree polynomial I have always seen it has no real root!

I am looking for a counterexample. please can anyone tell is there any general approach to solve this kind of problem? Often I do mistake in these.

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  • $\begingroup$ My understanding, which could be mistaken, is that if a polynomial has all real coefficients, then any complex roots come in pairs. Further, each of the complex pairs consists of a conjugate pair. That is, if one of the roots is $a+bi$ [where $\;a,b \,\in \mathbb{R}$], then $a-bi$ must also be one of the roots. $\endgroup$ Commented Jul 24, 2020 at 10:45
  • $\begingroup$ @user2661923 unable to go through ;;your comment $\endgroup$
    – user801681
    Commented Jul 24, 2020 at 10:48
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    $\begingroup$ Consider $f(x) = x^2 -x + 4,$ which is a polynomial with all real coefficients. One of the roots is $\frac{1}{2}\times \left(1 + i\sqrt{15}\right).$ Assuming the assertion in my previous comment is accurate, that would immediately imply that the conjugate $\frac{1}{2}\times \left(1 - i\sqrt{15}\right)$ is also a root of the polynomial. Again, assuming that my assertion is accurate, this will also be true if the polynomial is of 3rd degree, 4th degree, 5th degree ... $\endgroup$ Commented Jul 24, 2020 at 10:55

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For instance, $(x-1)(x-2)=x^2-3x+2$ has two real roots.

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  • $\begingroup$ Even simpler: $x^2$ ! (+1) $\endgroup$
    – Damien
    Commented Jul 24, 2020 at 10:40
  • $\begingroup$ totally agree. even i was also thinking that (x-1)(x+2)(x-3)(x-4) = 0 has 4 real roots although it's order is even. $\endgroup$
    – user801681
    Commented Jul 24, 2020 at 10:44

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