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Let $f(x):\mathbb{R}\to \mathbb{R}$ be a polynomial of degree $d$ with real coefficients. I was wondering if there exists any condition on the coefficients of this polynomial such that $f(x)=0$ has at least one complex root?

For $d=2$ this pertains to having negative discriminant. Is there any similar condition for $d=3$?

Thank you, in advance, for your response.

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    $\begingroup$ Just for the record: any polynomial with real coefficients, if it has one complex root, then its conjugate is also a root. Another useful result is Descartes' rule of signs. $\endgroup$ – rtybase Jan 30 at 22:22
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    $\begingroup$ Thanks for the comment. $\endgroup$ – Arthur Jan 30 at 22:37
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Of course if

$f(x) \in \Bbb R[x], \tag 1$

then complex roots come together in conjugate pairs, since if

$f(x) = \displaystyle \sum_0^n f_ix^i, \; f_i \in \Bbb R, \tag 2$

and

$f(\rho) = 0, \tag 3$

we have

$\displaystyle \sum_0^n f_i \bar \rho^i = \sum_0^n \bar f_i \bar \rho^i = \overline{\sum_0^n f_i \rho} = \overline{f(\rho)} = 0, \tag 4$

that is,

$f(\rho) = 0 \Longleftrightarrow f(\bar \rho) = 0. \tag 5$

As for the existence of such pairs of roots, this is a much more difficult issue; as pointed out, the case of quadratic $f$, $n = 2$, has been completely solved. In the case $n = 3$ that is

$f(x) = ax^3 + bx^2 + cx + d = 0, \tag 6$

there is in fact a known criterion for the existence of complex zeroes; indeed, if the discriminant

$\Delta = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2 < 0, \tag 7$

then (6) has a complex conjugate pair of roots. Of course, being of odd degree, $f(x)$ always has at least one real zero. See this wikipedia page on cubic polynomials for a detailed explanation. Similar reults are known in the quartic case $n = 4$, but the formulas are so complicated I won't copy them here; just click the link to get the full story.

Our general knowledge is exhausted by the cases $n \le 4$; for $n \ge 5$, one must resort to a variety of specialized methods.

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    $\begingroup$ Thank you for your comprehensive response! $\endgroup$ – Arthur Jan 30 at 23:05
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    $\begingroup$ @Arthur: my pleasure sir! And thanks for the "acceptance". $\endgroup$ – Robert Lewis Jan 30 at 23:06
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For $d=3$ there is at least one real root. Use polynomial division then solve the quadratic.

For $d\ge5$ there isn't even a way to find the roots.

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  • $\begingroup$ Thanks for your response. $\endgroup$ – Arthur Jan 30 at 22:38
  • $\begingroup$ "there isn't even a way to find the roots" is highly misleading. The roots can be approximated numerically to arbitrary accuracy. There is no formula by radicals for the roots, but such a formula is not particularly relevant to determining whether the roots are real. $\endgroup$ – Eric Wofsey Jan 31 at 4:27
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There is also a discriminant for cubics. For simplicity, assume that the cubic is monic with roots alpha,beta, gamma.

Then $D(f)=(\alpha-\beta)^2(\alpha-\gamma)^2(\beta-\gamma)^2$.

This is called the discriminant of a cubic. Using vieta's formulae you can write that using the coefficients of $f(x)=x^3+ax^2+bx+c$.

You get that $D(f)=a^2b^2+ 18abc − 4b^3− 4a^3c − 27c^2$.

If you have a repeated root, D=0. You can't have just one complex root, as they come in pairs. You have 2 complex roots iff D<0. You have 3 real roots iff D>0.

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  • $\begingroup$ Thanks for your response. $\endgroup$ – Arthur Jan 30 at 23:05

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