# Do cubics always have one real root? [closed]

I've seen a few conflicting pieces of information online.

So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients?

At very least could you give me a counterexample? A cubic with no real roots.

## closed as off-topic by user21820, Alex Provost, Eevee Trainer, Carl Mummert, Parcly TaxelMar 11 at 3:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Alex Provost, Eevee Trainer, Carl Mummert, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.

• It's more clear to say "at least one real root" or "a real root". There are real cubics that have three real roots (there are none that have exactly two though). – quid Mar 10 at 14:47
• @quid: If you count with multiplicity, that's true. If you don't, then you can have $f(x) = (x-1)(x-2)^2$. – Kevin Mar 10 at 17:22
• Indeed @Kevin I should have made that explicity. – quid Mar 10 at 19:15

## 3 Answers

One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $$f(z) =(z-i)^3$$. If you want three distinct complex roots, do something like $$f(z) = (z-i)(z+i)(z-2i)$$.

As you already know, a cubic with real coefficients always has at least one real root, so there is no counterexample of a cubic with real coefficients with no real roots.

A cubic with complex coefficients with no real roots is easy to find; take $$x^3+i$$.

• @Glen_b It is reasonable to say has one X to mean there exists an X, though it would be more clear to say there is at least one X. – jgon Mar 12 at 20:21
• @Glen_b I have edited to remove the ambiguity. – Servaes Mar 12 at 20:22

Over the complex numbers, every polynomial factors into roots. So we can take any cubic and write it as $$a(x-u)(x-v)(x-w)$$ where $$u, v, w$$ are the roots (they don't need to be distict) and $$a$$ is the leading coefficient. This lets us form polynomials with only complex roots such as $$(x-i)^3$$.

However, if all the original coefficients of the polynomial are real, and $$c$$ is a complex root, then its conjugate $$\bar{c}$$ must also be a root: complex roots to polynomials with real coeffecients must come in pairs. This is called the complex conjugate root theorem.

This means that a polynomial with real coefficients and odd degree will always have at least one real root, which answers the case for cubics. A quadratic with negative discriminant on the other hand has two, conjugate complex roots.