According to the theory of real polynomials, if complex roots occur; then they must occur in conjugate pairs. I want to ask if its proof and applicability to odd degree real polynomials, i.e. which definitely have no chance to have all complex roots (provided all, or odd number of roots are complex) in conjugate pairs.
I have a particular example where a real polynomial is of of cubic form, and the polynomial breaks into a product of a linear and a quadratic term. The quadratic term breaks further into 2 complex conjugate roots, while the linear part has only one complex root as the answer.
So, does the theory of complex roots occurring in conjugate pairs belongs only to even degree real polynomials.
Best Answer
The proof is pretty easy. Let $p(x) = a_n x^n + \cdots a_1 x + a_0$ be a real polynomial. We claim that complex conjugation plays nice with $p$. Let $\alpha \in \Bbb C$ be arbitrary. $$ \overline{p(\alpha)} = \overline{a_n \alpha^n + \cdots + a_1 \alpha + a_0} = \overline{a_n} \overline{\alpha^n} + \cdots \overline{a_1} \overline{\alpha} + \overline{a_0} $$
Since $a_i$ is real, this is just $$ = a_n \overline{\alpha^n} + \cdots a_1 \overline{\alpha} + a_0 = p(\overline{\alpha}) $$
So $\overline{p(\alpha)} = p(\overline{\alpha})$. What does this mean when $\alpha$ is a root?