From Aluffi, Algebra: Chapter $0$
If $s$ is any element of a ring $S$, then there is a unique ring homomorphism $\mathbb Z[x] \to S$ sending $x$ to $s$ and ‘extending’ the unique ring homomorphism $\iota : \mathbb Z \to S$. In this case the commutativity of $S$ is immaterial (why?).
The unique homomorphism $f: \mathbb Z[x] \to S$, where is $x \mapsto s$, is defined by mapping $f(a_0+a_1x+\cdots+a_nx^n)=\iota(a_0)+\iota(a_1)s+\cdots +\iota(a_n)s^n$ where $\iota: \mathbb Z \to S$ is the unique ring homomorphism.
Consider $(a_0 +a_2x^2)$ and $(b_1x+b_2x^2)$. Then $(a_0 +a_2x^2)(b_1x+b_2x^2)=a_0b_1x+a_0b_2x^2+a_2b_1 x^3+a_2b_2x^4$.
So,
$$f(a_0 +a_2x^2)=\iota(a_0) +\iota(a_2)s^2,$$
$$f(b_1x +b_2x^2)=\iota(b_1)s +\iota(b_2)s^2,$$
$$f(a_0b_1x+a_0b_2x^2+a_2b_1 x^3+a_2b_2x^4)=\iota(a_0b_1)s+\iota(a_0b_2)s^2+\iota(a_2b_1)s^3+\iota(a_2b_2)s^4.$$
However,
\begin{align*}
f(a_0 +a_2x^2)f(b_1x +b_2x^2)&=(\iota(a_0) +\iota(a_2)s^2)(\iota(b_1)s +\iota(b_2)s^2)\\
&=\iota(a_0)\iota(b_1)s+\iota(a_0)\iota(b_2)s^2+\iota(a_2)s^2\iota(b_1)s+\iota(a_2)s^2\iota(b_2)s^2.
\end{align*}
If $S$ does not commute then $f$ is not a ring hom? So, why does it not matter if $S$ is commutative?
Best Answer
The image of $\iota$ is automatically contained in the center of $S$. For example, $\iota(3) = 1_S + 1_S + 1_S$, so for any $x \in S$, we have $$\iota(3) x = (1_S + 1_S + 1_S) x = 1_S x + 1_S x + 1_S x = x + x + x = \cdots = x \iota(3).$$ (If you have not seen this result before, I will leave the formalization of the proof to you, along with the case of $\iota(n)$ for $n < 0$.)
Thus, for instance, $\iota(a_2) s^2 \iota(b_1) s = \iota(a_2) \iota(b_1) s^2 s = \iota(a_2 b_1) s^3$.