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

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 . It’s easy to see that and , 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 , then it will also have the zero .

At first, it may seem that the polynomial above, , is a counterexample. But this assumes that all irrational numbers can be written in the form . This is a gross mistake.

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but we also know the sum of roots that is -2 here

so when do irrational roots go in pairs?

According to the blog, irrational roots go in pair when one of the irrational roots can be expressed in the form a+\sqrt b

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