I can "see" it intuitively, though I do not know how correct this is: in a complex conjugate we change the sign of all imaginary parts, and since the effect of all imaginary parts cancels out on the whole, this change of sign would not matter.
I have tried, but I am unable to prove it. I tried using the polar forms.
Best Answer
For any $p=a_0+\cdots+a_nx^n\in\mathbb{C}[x]$, there is a conjugate polynomial $\overline{p}=\overline{a_0}+\cdots+\overline{a_n}x^n$. Note that if $z$ is a root of $p$, i.e. $$p(z)=a_0+\cdots+a_nz^n=0,$$ then $\overline{z}$ is a root of $\overline{p}$, i.e. $$\overline{p}(\overline{z})=\overline{a_0}+\cdots+\overline{a_n}\,\,(\overline{z})^n=0,$$ because $$\overline{p}(\overline{z})=\overline{a_0}+\cdots+\overline{a_n}\,\,(\overline{z})^n=\overline{a_0}+\cdots+\overline{a_nz^n}=\overline{a_0+\cdots+a_nz^n}=\overline{0}=0,$$ where we have used that $\overline{ab}=\overline{a}\,\overline{b}$ and $\overline{a+b}=\overline{a}+\overline{b}$ for all $a,b\in\mathbb{C}$.
Now it only remains to note that if $p\in\mathbb{R}[x]$, then $p=\overline{p}$.