Let $f(x) \in \mathbb{R}[x]$ with $\deg f=2$. Show that $f(x)$ is irreducible over $\mathbb{R}$ if and only if $f(x)=(x-a)^2 +b^2$, where $a,b \in \mathbb{R}$ and $b \neq 0$.
I don't know how to do the first part of this exercise and I need some help.
For the way back I did:
If $f(x)=(x-a)^2 +b^2$, to get $f(x)=0$, We need $a=0$ and $b=0$. But, by hypothesis, $b \neq 0$. So $f(x)$ doesn't have roots in $\mathbb{R}$, therefore $f(x)$ is irreducible.
Is this part right?
Hint:
If $f(x)$ is irreducible, it has two conjugate roots $\zeta,\bar\zeta$, so it factors over $\bf C$ as \begin{align} f(x)&=(x-\zeta)(x-\bar\zeta)=x^2-(\zeta+\bar\zeta)x+\zeta\bar\zeta \\ &=x^2-2\operatorname{Re}(\zeta)x+|\zeta|^2=\bigl(x-\operatorname{Re}(\zeta)\bigr)^2-\bigl(\operatorname{Re}(\zeta)\bigr)^2+|\zeta|^2 \\ &=\cdots \end{align}