# How many complex roots does a cubic equation have?

Aug 10, 2015

Since every real number can be thought of as a complex number, one way to answer this is to say the answer is 3, after "counting multiplicities". On the other hand, if you are after non-real complex roots, and if the coefficients of your cubic equation are real numbers, the answer will be 0 or 2.

#### Explanation:

A real number $a$ can be thought of as the complex number $a + 0 i$. A cubic such as $\left(x - 2\right) {\left(x - 5\right)}^{2} = {x}^{3} - 12 {x}^{2} + 45 x - 50$ would be an example that would then have three "complex" roots. The number 2 would be a root of multiplicity 1 and the number 5 would be a root of multiplicity 2.

If you are after non-real complex roots in a situation with real number coefficients, the same example above would provide a situation where there are 0 complex roots. Examples where there are two non-real complex roots are situations where they occur in complex conjugate pairs . For example, $\left(x - \left(3 + 4 i\right)\right) \left(x - \left(3 - 4 i\right)\right) \left(x + 3\right) = {x}^{3} - 3 {x}^{2} + 7 x + 75$ has roots $x = - 3$, $x = 3 \setminus \pm 4 i$, and the complex numbers $3 + 4 i$ and $3 - 4 i$ are called complex conjugates.

It can be proved that if a polynomial has real coefficients, then the non-real complex roots of such a polynomial, if there are any, must occur in these complex conjugate pairs. That's why there can't be such cubics with 1 or 3 non-real complex roots.

One other thing worth mentioning: by the Intermediate Value Theorem, all cubics with real coefficients must have at least 1 real root (giving another reason such a cubic can't have 3 non-real complex roots).