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: $$x^3-x^2-x-\frac{5}{27}=0$$

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.
