Do Irrational Roots Come in Pairs? (Part 1)

The short answer is NO.

This past week I had a conversation with some of my colleagues about this question. A long time ago I put this on my list of math topics I needed to think about more. So I was grateful for an opportunity to think through this problem. The polynomial we were working with was

$f(x)=-x^5+2x^4+7x^3+x^2-4x+1$

We were talking about Rational Root Theorem and Descartes’ Rule of Signs. Descartes’ tells us that there are 3 or 1 positive real zeros and 2 or 0 negative real zeros. Rational Root Theorem tells us that if there any rational roots they will be $\pm1$ . It’s easy to see that $f(1)=6$ and $f(-1)=2$ , so neither of these are zeros. If we look at the graph, it crosses the x-axis only once. That means that this fifth degree polynomial has 4 nonreal roots and 1 irrational root. How is this possible? Don’t irrational roots come in pairs?

The Irrational Root Theorem

Here’s the irrational root theorem (in my own words):

If a polynomial with rational coefficients has a zero of the form $x=a+b\sqrt{c}$ , then it will also have the zero $x=a-b\sqrt{c}$ .

At first, it may seem that the polynomial above, $f(x)$ , is a counterexample. But this assumes that all irrational numbers can be written in the form $a+b\sqrt{c}$ . This is a gross mistake.