# Condition for Having Complex Roots in Polynomial Equations

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.

• 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. – rtybase Jan 30 at 22:22
• Thanks for the comment. – 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

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

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.

• Thank you for your comprehensive response! – Arthur Jan 30 at 23:05
• @Arthur: my pleasure sir! And thanks for the "acceptance". – 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.

• Thanks for your response. – Arthur Jan 30 at 22:38
• "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. – 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.

• Thanks for your response. – Arthur Jan 30 at 23:05