Non unique factorization of integer valued polynomials

This can even be done with one variable: $$ 2\cdot \left(\frac{x(x+1)}{2}\right)=\big(x\big)\cdot\big(x+1\big). $$ If you prefer to avoid irreducibles that become units in $\mathbb{Q}$: $$ \left(\frac{x(x+1)}{2}\right)\cdot\left(\frac{(x+2)(x+3)}{2}\right) = \left(\frac{x(x+3)}{2}\right)\cdot\left(\frac{(x+1)(x+2)}{2}\right). $$

Worth emphasis is how extremely different the lengths of irreducible factorizations can be in such nonunique factorizations in the ring of integer-valued polynomials. For example

$$ n {x\choose n}\, =\, (x-n+1){x\choose n-1}$$

The RHS is a product of two irreducibles, but the LHS can have an arbitrarily large number of irreducible factors by choosing $n$ with many prime factors.

For completeness, below is a proof of the irreducibility of $x\choose n$ from the 2016 Monthly paper What You Should Know About Integer-Valued Polynomials by Cahen and Chabert.