Key role of distributive law in expansion and factorization (from Spivak's calculus) 
Hi, I have a basic question concerning definition of the word 'factorization'. Does Spivak consider factorization as development of factors ? He goes from saying the "factorization" $x^2-3x+2=(x-1)(x-2)$ is really a triple use of $P9$ and goes on showing development.
$P9$ says : If $a,b,$ and $c$ are any numbers, then :
$a \cdot (b+c)= a \cdot b  +a \cdot c$
Also, when Spivak does the following : $(x-1)(x-2)=x(x-2)+(-1)(x-2)$ does he use any property or just assumes it as like this ? I know whats happening, just curious if there's any justification to it.
Thank you ! 
 A: I can see how a reader might easily be confused by the passage that you quoted. It seems that Spivak's goal here is to emphasize the fundamental role played by $\rm P9$ (the Distributive Law) by highlighting the key that it plays in even the simplest of arithmetical (ring-theoretic) calculations, such as factoring a quadratic polynomial, law of signs, etc. Because the distributive law is the only law that connects the additive structure to the multiplicative structure in a ring, it must be used in any proof of a theorem that is not purely additive, or purely multiplicative.
Without the distributive law a ring degenerates to a set with two completely unrelated additive and multiplicative structures. So, in a sense, the distributive law is a keystone of the ring structure.
The proof that he gives of the "factorization" $x^2-3x+2 = (x-1)(x-2)$ is not meant to be read as the derivation (discovery) of the factorization but, rather, simply as a proof that this equation is true. The point is simply to illustrate the key role that $\rm P9$ plays in proving this equality.
Normally transforming the lhs to rhs, i.e $\,x^2-3x+2$ to $(x-1)(x-2),$ is called factorization, and the reverse direction is called expansion. Each necessarily involves multiple applications of the distributive law $\rm P9$ (in various directions). Probably the reason Spivak put scare quotes around the word "factorization" is because the equation could be read either way, as a factorization or an expansion. What makes matters more confusing is that his proof of the "factorization" equality actually goes from rhs to lhs, so it would normally by called an expansion of $\,(\color{#c00}{x-1})(x-2)\,$ by distributing the factor $\,x-2\,$ into the summands $\,\color{#c00}x\,$ and $\,\color{#c00}{-1},\,$ yielding
$$\begin{align} &(\color{#c00}{x}\ +\ (\color{#c00}{-1}))\, (x-2) \\[.4em] =\ & \color{#c00}x(\color{#0a0}{x-2}) + (\color{#c00}{-1})(x-2)\end{align}$$
then distribute $\,\color{#c00}x\,$ into the summands $\,\color{#0a0}x\,$ and $\,\color{#0a0}{-2},\,$ and similarly distribute $\color{#c00}{-1}$ yielding
$$x(\color{#0a0}{x}) + x(\color{#0a0}{-2}) + (-1)x + (-1)(-2)$$
Thus this complete expansion applied the distributive law four times in total.
In any case, don't let the unoriented terminology disorient you.  You can safely ignore this, since Spivak's point is not to give a definition of factorization or expansion but, rather, to show prototypical applications of $\rm P9$.
A: Factorization makes sens in any ring which is basically a set equipped with $+,-,\cdot$ operations that satisfy the usual conditions that we use for calculations.
One such condition is distributivity, i.e. $+$ and $\cdot$ together must satisfy
$$a\cdot(b+c)=a\cdot b+a\cdot c$$
for all elements $a,b,c$.
There are several rings. For example, the set of integer numbers, $\Bbb Z$ (with the usual $+,-,\cdot$ operations) is a ring, and factorization $a=bc$ here has the original meaning.
The other ring in question is the ring of polynomials (denoted by $R[x]$) with coefficients from a ring $R$ (which is now either $\Bbb Z$, $\Bbb Q$ or $\Bbb R$, doesn't really matter). So, elements of $R[x]$ are formal finite sums of terms $r_nx^n$ where $r_n\in R$ and $x$ denotes the indeterminant. This also satisfies the usual properties (i.e. the ring axioms) -- e.g. multiplication is defined so that it satisfies distribution --, so it is also a ring, and thus, factorization in $R[x]$ makes sense, and we indeed have
$$(x-1)(x-2)=x^2-3x+2$$
by distributivity (P9) which also holds among polynomials.
