# How many complex roots does this polynomial $x^3-x^2-x-\frac{5}{27}=0$ have?

The fundamental theorem of algebra states that any polynomial to the $n^{th}$ will have $n$ roots (real and complex). But I know that complex roots only come in pairs because they are conjugates of each other. According to the theorem, $x^3-x^2-x-\frac{5}{27}=0$ should have 3 real and complex roots combined, in this polynomial $n=3$. If you graph this you can see that there are only 2 real roots, that means there must be 1 complex root so that the real and complex roots add up to 3, but we know that a polynomial cannot have an odd number of complex roots. So can someone explain this to me? Am I missing something, how many real and how many complex roots does this polynomial have: $$x^3-x^2-x-\frac{5}{27}=0$$

Counting repeated roots, it will, of course, be $3$.

To find out how many roots are repeated roots: If $p(x)$ has repeated roots, they are also roots of $p'(x)$ of one less degree. So compute the GCD of $p(x)$ and $p'(x)$ and subtract the degree of the result from the degree of $p(x)$ to get the number of unique roots of $p(x)$.

So:

$$\deg p(x) - \deg \gcd(p(x),p'(x))$$

Is the value you seek for unique roots.

In this case, you have a repeated root.

The function factors as

$$f(x)=\frac1{27}(3x-5)(3x+1)^2.$$

It has 3 real roots: $5/3, -1/3$ and again $-1/3$, so only 2 of them are distinct. This is why you see only 2. You could have also seen that there is a double zero without factoring it, because non-real complex roots come in conjugate pairs.