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

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?

• In the title, you seem to have a typo (K instead of R) Commented Oct 13, 2019 at 11:24
• If the discriminant is non-negative, we have at least one real root $r$ , hence the polynomial is reducible ($x-r$ is a factor) Commented Oct 13, 2019 at 11:25
• @Peter thanks, I've fixed it! Commented Oct 13, 2019 at 11:27
• At the bottom of the question, it should be R as well , right ? Commented Oct 13, 2019 at 11:28
• Yeah, sorry ... Commented Oct 13, 2019 at 11:40

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