Try replacing $x$ with $u^2$ to clear the fractional exponents.
If $u^2=x$, we get $3x^{3/2}-9x^{1/2}+6x^{-1/2} = 3(u^2)^{3/2}-9(u^2)^{1/2}+6(u^2)^{-1/2}$
$= 3u^3-9u+6u^{-1}$
Next, factor out $u^{-1}$ and get $3u^{-1}(u^4-3u^2+2)$.
This is now quadratic in $u^2$ (which is $x$) so we get $3x^{-1/2}(x^2-3x+2) = $
$$ \frac{3(x-2)(x-1)}{\sqrt{x}}$$
Of course, you can avoid introducing $u$ if you see that factoring out $x^{-1/2}$ at the beginning leaves you with a quadratic in $x$.
Your first term, upon division: $x^2$ is correct $\large\checkmark$ (in the quotient), leaving
$$ x - 2 | -x^2 - 4x + 12\quad\large\checkmark$$
I know that the leading term goes into the inner leading term $-x$ times or however you say that.
$$ x - 2 | - 2x + 12 \quad \longleftarrow \text{error}$$
Here you subtracted incorrectly: We should have $x^2 - x$ in the quotient, that's correct, but multiplying $-x(x - 2) = -x^2 + 2x$
So when we subtract, we subtract, from $(-x^2 - 4x + 12) - (-x^2 + 2x) = -6x + 12$.
Now, we have $$ x -2 \mid -6x + 12$$ and so our ongoing quotient becomes $x^2 - x {\bf - 6}$
which, leaves a zero remainder since $-6(x-2) = -6x + 12$, as desired.
So...we have that $$\frac{x^3 - 3x^2 -4x + 12}{x - 2} = x^2 - x - 6$$
Or, that is, $$x^3 - 3x^2 - 4x + 12 = (x -2)(x^2 - x - 6) = (x-2)(x +2)(x - 3)$$
Best Answer
No. I will consider the case of factorizing over the complex numbers, or any sub-field (e.g. the real numbers). Most of my discussion will not really depend on the choice of field.
Any factorization (into polynomials) must be into polynomials of degree $1$, i.e. $$ a^2 + b^2 + c^2 \overset?= (\alpha_1 a + \beta_1 b + \gamma_1 c + \delta_1)(\alpha_2 a + \beta_2 b + \gamma_2 c + \delta_2) $$ where $\alpha_1,\dots,\delta_2$ are complex numbers. If you don't mind doing some work, you can expand this out. You might as well rescale $\alpha_1 \leadsto \alpha_1' = 1$, $\beta_1 \leadsto \beta_1' = \beta_1 / \alpha_1$, $\gamma_1 \leadsto \gamma_1' = \gamma_1 / \alpha_1$, $\delta_1 \leadsto \delta_1' = \delta_1 / \alpha_1$, $\alpha_2 \leadsto \alpha_2' = \alpha_1\alpha_2$, ..., $\delta_2 \leadsto \delta_2' = \delta_2\alpha_2$. This lets you assume that $\alpha_1 = 1$, and you immediately conclude $\alpha_2 = 1$. Now looking at the coefficients of $ab$, you see that $\beta_1 = -\beta_2$, and on the other hand the coefficient of $\beta^2$ gives $\beta_1\beta_2 = 1$; from these equations you can conclude that $\beta_1 = \pm i$, and $\beta_2$ is its negation. A similar argument looking at the coefficients of $ac$ and $c^2$ gives $\gamma_1 = \pm i$, and $\gamma_2$ is its negation. But this leads to a contradiction when looking at the coefficient of $bc$.
Let me mention an important trick when looking for factorizations of highly symmetric things. It is a (nontrivial) theorem that factorization of polynomials (in any number of variables) is unique in the following sense: if you have two factorizations of the same polynomial, such that each factor itself has no factors, then the two factorizations agree up to reordering the list and multiplying terms on the list by nonzero (complex, say) numbers.
So, let's assume we have a factorization, say the one above. We can try to generate other factorizations. Well, the left hand side is real, in the sense of being invariant under complex conjugation. Thus the complex conjugate of the right hand side is also a factorization. It follows that we have either one of two possibilities:
Possibility 1: The lists are not reordered. Therefore, there is a nonzero complex number $\eta$ such that $\bar\alpha_1 = \eta \alpha_1$, $\bar \beta_1 = \eta\beta_1$, $\bar \gamma_1 = \eta\gamma_1$, $\bar \delta_1 = \eta\delta_1$, $\bar\alpha_2 = \eta^{-1} \alpha_2$, $\bar \beta_2 = \eta^{-1}\beta_2$, $\bar \gamma_2 = \eta^{-1}\gamma_2$, $\bar \delta_2 = \eta^{-1}\delta_2$. It follows that $\bar\eta = \eta^{-1}$. Upon multiplying the first factor by $\sqrt{\eta}$ and the second by $\sqrt{\eta}^{-1}$, we can assume without loss of generality that $\eta = 1$. But then both factors are real, and this is impossible: the LHS has no zeros except for $\alpha = \beta = \gamma = 0$, whereas each factor on the right has a two-dimensional space worth of zeros.
Possibility 2: Under complex conjugation, the two terms switch places. Again, we can multiply by a constant, and thereby assume that $\alpha_2 = \bar \alpha_1$, $\beta_2 = \bar \beta_1$, etc.
What other symmetries are there? Well, the left-hand side is invariant under each of $a \mapsto -a$, $b \mapsto -b$, and $c\mapsto -c$. Up to some casework, this gives linear relations between the coefficients. The left-hand side is also invariant under switching $a$ with $b$, say.
Using all of these symmetries dramatically reduces the possible space of factorizations, as any factorization must respect all symmetries up to some scalar factors.
Here's one final way to think about factorization problems. Let's fix a non-zero value for $c$, and consider the space of solutions $(a,b)$ to $0 = a^2 + b^2 + c^2$ ($a$ and $b$ may be complex numbers). If $c^2$ were negative and we were looking for real solutions, this would be a circle. Your ability to visualize in $\mathbb C^2$ may not be great, but I assure you that the space of solutions is still something like a circle. In particular, it is not a linear subspace of $\mathbb C^2$.
On the other hand, the space of solutions $(a,b)$ to any equation of the form $0 = p(a,b)q(a,b)$ where $p,q$ are degree-$1$ is a union of two (usually intersecting) linear subspaces in $\mathbb C^2$.
This type of argument also shows that $a^2 + b^2 + c^2$ has no factorizations over the complex numbers.
Finally, a commenter suggested looking at the quaternions. The quaternions have lots of behaviors that are different from commutative fields like $\mathbb C$. Indeed, over the quaternions, we have $$ (ai + bj + ck)(-ai - bj - ck) = a^2 + b^2 + c^2 $$ provided $a,b,c$ are real (and hence commute with everything). If $a,b,c$ are allowed also to be quaternionic, then I don't see any convenient factorization, but I don't have a proof that it is impossible.