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.

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ Thanks for the comment. $\endgroup$ – Arthur Jan 30 at 22:37

Of course if

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

then complex roots come together in conjugate pairs, since if

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


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

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.

  • 1
    $\begingroup$ Thank you for your comprehensive response! $\endgroup$ – Arthur Jan 30 at 23:05
  • 1
    $\begingroup$ @Arthur: my pleasure sir! And thanks for the "acceptance". $\endgroup$ – Robert Lewis Jan 30 at 23:06

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.

  • $\begingroup$ 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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.