Prove, that $p\in\mathbb{R}[x]$ with $p=x^2+ax+b$ is irreducible if and only if $a^2<4b$.

Question

Prove, that $p\in\mathbb{R}[x]$ with $p=x^2+ax+b$ is irreducible if and only if $a^2<4b$.

$"\leftarrow"$ Let $a^2<4b \iff b>\frac{a^2} 4$

\begin{align} &x^2+ax+b=0\\ x_{1/2}=&\frac a 2\pm \underbrace{\sqrt{\frac{a^2}{4}-b}}_{\text{is negative: } b>\frac{a^2} 4} \end{align} $\implies$ We have no roots! Therefore, the polynomial is irreducible

$"\rightarrow"$ Assume, $p$ is irreducible, $\nexists q,z\in\mathbb{R}[x]:p=q\cdot z$. How to go on now?

Answer 1

For $\rightarrow$ you can prove the counterpositive. Assume that $a^2\geq 4b$ and conclude from that that $p$ factors.

The same quadratic formula gives you that the factorization works over the reals, since this time the radicand is non-negative.

Answer 2

For a clean proof, you can do the following: observe that $x^2+ax+b$ is reducible exactly when it can be written into the product of linear factors $(x-\alpha_1)(x-\alpha_2)$, which is true exactly when it has two real roots. This is true exactly when the discriminant is strictly positive. This does both directions at once!

Caveat: We do need to know the fact that we can write $x^2+ax+b=(x-\alpha_1)(x-\alpha_2)$ exactly when it has two real roots. The forward direction is trivial, but the other direction requires slightly more thought. It follows from the fact that $\mathbb R[x]$ is a Euclidean domain, as it allows us to use the Remainder Theorem (equivalently, polynomial division).