# How to deal problems in finding real roots of a polynomial?

After encountering many numerical problems related to finding real roots of a polynomial, I have fixed a simple path:

1. For an odd-degree polynomial: complex roots come in pairs. If a polynomial is of degree $$d$$ which is odd then the polynomial must have at least one real root.

2. For a even-degree polynomial I have always seen it has no real root!

I am looking for a counterexample. please can anyone tell is there any general approach to solve this kind of problem? Often I do mistake in these.

• My understanding, which could be mistaken, is that if a polynomial has all real coefficients, then any complex roots come in pairs. Further, each of the complex pairs consists of a conjugate pair. That is, if one of the roots is $a+bi$ [where $\;a,b \,\in \mathbb{R}$], then $a-bi$ must also be one of the roots. Commented Jul 24, 2020 at 10:45
• @user2661923 unable to go through ;;your comment
– user801681
Commented Jul 24, 2020 at 10:48
• Consider $f(x) = x^2 -x + 4,$ which is a polynomial with all real coefficients. One of the roots is $\frac{1}{2}\times \left(1 + i\sqrt{15}\right).$ Assuming the assertion in my previous comment is accurate, that would immediately imply that the conjugate $\frac{1}{2}\times \left(1 - i\sqrt{15}\right)$ is also a root of the polynomial. Again, assuming that my assertion is accurate, this will also be true if the polynomial is of 3rd degree, 4th degree, 5th degree ... Commented Jul 24, 2020 at 10:55

For instance, $$(x-1)(x-2)=x^2-3x+2$$ has two real roots.
• Even simpler: $x^2$ ! (+1) Commented Jul 24, 2020 at 10:40