Let $F : \mathbb{R}[x] \to \mathbb{C}$ be given by $F( f(x) ) = f(i) $
Want to show $\ker F = \langle x^2 + 1 \rangle$ (ideal generated by $x^2+1$) .
MY attempt: If $f \in \langle x^2 + 1 \rangle $, then can write $f(x) = (x^2 + 1) g(x) $ for some $g \in \mathbb{R}[x] $. Notice $F ( f ) = f(i) = (i^2 +1)g = 0 $. Hence $f \in \ker F $. So $\langle x^2 + 1 \rangle \subset \ker F $.
Next, if $f(x) \in \ker F $, then $F(f) = f(i) = 0 $. So we can write $f(x) = (x-i)g(x) $. the conjugate is also a root so $f(-i) = (-2i)g(-i) = 0 $ and so $g(-i) = 0$ hence can write g(x) = (x+i)h(x) for some $h(x) \in \mathbb{R}[x]$. So,
$$ f(x) = (x-i)(x+i)h(x) = (x^2+1) h(x)$$
Clearly $h(x) \in \mathbb{R}[x]$ hence $f \in \langle x^2 +1 \rangle $. And so $\ker F = \langle x^2 +1 \rangle $.
is this correct solution?
The proof is okay but you should remark that you are using the fact that for a polynomial $p\in\mathbb R[x], p(z) = 0 \Leftrightarrow p(\bar z) = 0$. This should be made a bit more explicit.
Note that you could use a weaker statement as well: $x^2 + 1$ is irreducible in $\mathbb R[x]$ and $i^2 + 1 = 0$. This implies that if $p(i)=0$ for some $p\in\mathbb R[x]$ then necessarily $x^2+1 | p$.