Irreducible and prime elements at \mathbb{R}

Question

I found a lots of materials about this problem, but can someone explain a elementary way? I cannot understand the abstract algebra thing..

For all $F(x) \in \mathbb{R}[x]$ and $F(X)$ is irreducible, then for all $G,H \in \mathbb{R}[x]$ such that $F|GH$, $F|G$ or $F|H$ holds?

Answer 1

Suppose $F\mid GH$, but $F$ does not divide $G$. Since $F$ is irreducible, this implies the greatest common divisor between $F$ and $G$ is 1. So (by the Euclidean Algorithm) there exists polynomials $P,Q$ such that $PF + QG = 1$. Multiplying this equation by $H$ gives $PFH+QGH=H$. Since $F$ divides both terms on the left side of the equation, $F$ also divides $H$.