Contradiction to Irrational Roots Come in Conjugate Pairs?

During class, we discussed a polynomial while learning about the upper bound theorem and Descarte's Rule of Signs:

2x^3 + x^2 - 8x - 3

Upon applying many theorems to analyze this polynomial and graphing it as well, we discovered that there are two negative real roots and one positive real root.

Using the rational root theorem, we discovered that none of the three real roots are rational because they don't match any of the 1,3 / 1,2 possible rational roots from the rational root theorem.

However, this means there are three irrational roots. That's the problem, we can't figure out why this is possible, because mathematical theorems say that irrational roots come in conjugate pairs.

Can anybody explain why this is possible, or whether there is some flaw in the irrational root theorem? Thanks.

Re: Contradiction to Irrational Roots Come in Conjugate Pairs?

The roots of cubic equations often involve cube roots, rather than square roots. (And often there are square roots inside the cube roots.) So the irrational conjugate root theorem does not apply.