Is this proposition about polynomials obvious or not?

Question

I am reading an algebra book now.

Let $p(x) \in \mathbb{R}[x]$ and $\deg p = n$.
Let $\alpha \in \mathbb{C} - \mathbb{R}$ and $p(\alpha) = 0$.

Then, $p(\bar{\alpha}) = \overline{p(\alpha)} = 0$.

So, $p(x)$ is divisible by $(x-\alpha)(x-\bar \alpha)$.

$(x-\alpha)(x-\bar \alpha) = x^2 - 2 \Re(\alpha) x + |\alpha|^2 \in \mathbb{R}[x]$.

So, there is a polynomial $q(x) \in \mathbb{R}[x]$ such that $p(x) = (x^2 - 2 \Re(\alpha) x + |\alpha|^2) q(x)$.

Why $q(x) \in \mathbb{R}[x]$?

Is it necessary to prove that $q(x) \in \mathbb{R}[x]$?
Or,
is it obvious that $q(x) \in \mathbb{R}[x]$?

My proof is here:

There is $q_1(x) \in \mathbb{C}[x]$ such that $p(x) = (x-\alpha)(x-\bar \alpha) q_1(x)$ in $\mathbb{C}[x]$.
There is $q_2(x), r_2(x) \in \mathbb{R}[x]$ such that $p(x) = (x^2 - 2 \Re(\alpha) x + |\alpha|^2) q_2(x) + r_2(x)$ in $\mathbb{R}[x]$ and $2 > \deg r_2$.
$p(x) = (x-\alpha)(x-\bar \alpha) q_1(x) = (x^2 - 2 \Re(\alpha) x + |\alpha|^2) q_2(x) + r_2(x)$ in $\mathbb{C}[x]$.
By a theorem about long division, $q_1(x) = q_2(x), r_2(x) = 0$.
So, $p(x) = (x^2 - 2 \Re(\alpha) x + |\alpha|^2) q_2(x)$, and $q_2(x) \in \mathbb{R}[x]$.

Answer 1

The key point is that if $f,g$ are nonzero polynomials, all of $f$s coefficients are real, and $g$ has some non-real coefficient, then their product $f\cdot g$ also has some non-real coefficient. This is easy to prove - think about the lowest-degree term of $g$ with non-real coefficient ...

Now consider $f(x)=x^2-2\mathfrak{R}(\alpha)x+\vert\alpha\vert^2$ and $g(x)=q(x)$, and note that $p=f\cdot g$ has all-real coefficients ...

Answer 2

r(x) = (x-a)(x-$\bar{a}$) $\in$ $\mathbb{R}[x]$. Since p(x) $\in$ $\mathbb{R}[x]$ and r(x) | p(x), there exists a polynomial q(x) such that p(x) = q(x)r(x). q(x) could not have imaginary coefficients because after multiplication p would have to have complex coefficients as well.