Confusion about the expansion of $\;x(2x - 3)(3x + 4)$ For this algebraic product: 
$\;x(2x - 3)(3x + 4)$
What I expected to do is:
$$\begin{align} x(2x - 3)(3x + 4) & = 2x^2 -3x(3x + 4) \\ & = 2x^2 -9x^2 -12x \\ & = -7x^2 -12x\end{align}$$
However, what the book I'm reading does is:
$$\begin{align} x(2x - 3)(3x + 4) & = (2x^2 - 3x)(3x + 4) \\ &= 6x^3 +8x^2 -9x^2 -12x \\ & = 6x^3 -x^2 -12x\end{align}$$
So, in this part: $x(2x - 3)$, why the product become between parenthesis??

First off, to address a misunderstanding expressed in your comment:
$$a - b(c + d) \neq  (a - b)(c + d)$$ Rather $$(a-b)(c+d) = a(c+d) - b(c+d) = ac + ad -bc -bd$$
Now, we address the posted problem. If proceeding left to right, $x$ needs to multiply both $2x$ and $-3$, to give $(2x^2 - 3x)$, all of which needs to then multiply $(3x + 4)$. So think of first multiplying the two factors $x$ and $2x- 3$ to get $$x(2x - 3) = 2x^2 - 3x$$ So now, we have multiply this partial product with the factor $3x + 4$ to get $$\begin{align}(2x^2 - 3x)(3x + 4) & = 2x^2(3x + 4) - 3x(3x + 4) \\ \\ & = 6x^3 + 8x^2 - 9x^2 -12 x \\ \\ &= 6x^3 - x^2 - 12 x\end{align}$$
Another way to approach this is to expand the two right-most factors first, then multiply that expansion by $x$:
$$\begin{align} x(2x - 3)(3x + 4) & = x(2x\cdot 3x+  2x\cdot 4 -  3\cdot 3x - 3\cdot 4) \\ \\ &= x(6x^2 -x -12) \\ \\ & = 6x^3 - x^2 -12x\end{align}$$
