What is the essence of the constant factor in the standard definitions of the discriminant? Let $f(x) = x^m+\sum_{j=0}^{m-1}f_{m-j}x^j\in P[x]$ be a monic polynomial over a field $P$ and let $f(x) = (x-\alpha_1)\cdot\ldots\cdot(x-\alpha_m)$ be a factorization of $f$ over an extension field $Q$ of $P$.
Then it is quite natural to consider a value (called discriminant) $$\prod_{1\leq i<j\leq m}(\alpha_i-\alpha_j)^2,$$
whose being zero or not determines whether or not the polynomial has any multiple roots (in any extension field of $P$, not only $Q$).
But, when $f$ is not monic, say $f = f_0 \cdot x^m+\sum_{j=0}^{m-1}f_{m-j}x^j $, a usual definition of the discriminant reads 
$$
f_0^{2m-2}\cdot \prod_{1\leq i<j\leq m}(\alpha_i-\alpha_j)^2.$$
So, my question is: 
What is the essence of the factor $f_0^{2m-2}$?
 A: Perhaps you will find the following helpful, which expands a bit on Robert Israel's answer:  Write $f(x)=f_0x^m+\ldots+f_m=f_0\cdot(x-\alpha_1)\cdots(x-\alpha_m).$  Multiplying out the RHS we get $f(x)=f_0\cdot \sum_{j=0}^m(-1)^j\sigma_j(\alpha_1,\ldots,\alpha_m)\cdot x^{m-j}$ where $\sigma_j=\sigma_j(\alpha_1,\ldots,\alpha_m)$ is the $j^{th}$ elementary symmetric polynomial in $\alpha_1,\ldots,\alpha_m$.  Then matching coefficients we get that $$\frac{f_j}{f_0}=(-1)^{j}\cdot\sigma_j.$$
Now in the polynomial ring ${\mathbb{C}}[\alpha_1,\ldots,\alpha_m]$, the symmetric group $W=\mathfrak{S}_m$ acts by permuting variables, and a well known fact from invariant theory says that the invariant subring is generated by the elementary symmetric polynomials, i.e.
$${\mathbb{C}}[\alpha_1,\ldots,\alpha_m]^W={\mathbb{C}}[\sigma_1,\ldots,\sigma_m]={\mathbb{C}}\left[\frac{f_1}{f_0},\ldots,\frac{f_{m-1}}{f_0}\right].$$
Now the discriminant
$$\Delta^2=\prod_{1\leq i<j\leq m}(\alpha_i-\alpha_j)^2$$
is also invariant under $W$, hence we must have that 
$$\Delta^2=P\left(\frac{f_1}{f_{0}},\ldots,\frac{f_{m-1}}{f_0}\right)$$
for some polynomial $P\in{\mathbb{C}}[y_1,\ldots,y_m]$.  Multiplying $\Delta^2$ by an appropriate power of $f_0$ will clear denominators to give you an honest polynomial in the $f_0,\ldots,f_m$.  Certainly the exponent $m(m-1)$ will work, but it is not quite clear to me why $2(m-1)$ will, as you claim in your original statement...and perhaps this is the point of your question...?
A: A trivial observation, which was not pointed out so far: the non-normalized discriminant is 'better' in that it comes 'closer' to being a 'homomorphism' $P[x]\rightarrow P$, though it 'still' is not such a homomorphism, it is merely 'multiplicative up to a multiplicative computable polynomial-in-the-coefficients constant'. 
I think one could characterize the factor $([x^{\mathrm{deg}(f)}](f))^{2\deg(f)-2}$, that the OP is asking the 'essence' of, as the unique factor which gives the 'most efficient' multiplicativity-identity, though I have never seen this and don't have time to try to make this observation rigorous. I expect this to be widely known. It seems some sort of answer to the OP's question for the 'essence' of the normalizing factor, and perhaps even for the OP's question about the '$2m-2$'.
Here is the trivial observation.
Let $\mathrm{disc}$ denote the (standard) discriminant, in the sense of  the (corrected) OP.
Let $\mathrm{d}f$ denote the degree of a polynomial $f$.
Let $\mathrm{ndisc}(f) := \frac{1}{a_{\mathrm{d}f}^{2\mathrm{d}f-2}}\cdot \mathrm{disc}(f)$ denote the normalized discriminant.1 .
( $\mathrm{ndisc}$ is defined since a leading coefficient by definition is non-zero )
Then, for all polynomials $f_0,f_1$ of degree $\geq 1$ we have the useful and normalization-factor-free identity

$\mathrm{disc}(f_0\cdot f_1) = \mathrm{resultant}(f_0,f_1)\cdot \mathrm{disc}(f_0)\cdot \mathrm{disc}(f_1)$. ${}\qquad$ (multiplicative.up.to.a.polynomial.factor)

For the 'normalized discriminant' $\mathrm{ndisc}$, i.e. the one without the factor that the OP is asking about, we however find a less efficient identity. 
Using that $\mathrm{d}( f_0\cdot f_1) = \mathrm{d}f_0 + \mathrm{d}f_1$ and that the leading coefficient of $f_0\cdot f_1$ is $a_{\mathrm{d}f_0} \cdot a_{\mathrm{d}f_1}$, we find the less efficient identity
$\small\text{$(a_{\mathrm{d}f_0}\cdot a_{\mathrm{d}f_1})^{2\cdot(\mathrm{d}f_0+\mathrm{d}f_1)-2}\cdot \mathrm{ndisc}(f_0\cdot f_1) = \mathrm{resultant}(f_0,f_1)\cdot a_{\mathrm{d}f_0}^{2\mathrm{d}f_0-2}\cdot a_{\mathrm{d}f_1}^{2\mathrm{d}f_1-2}\cdot\mathrm{ndisc}(f_0)\cdot \mathrm{ndisc}(f_1)$ }$
equivalently

$\small\text{$\mathrm{ndisc}(f_0\cdot f_1) = \frac{\mathrm{resultant}(f_0,f_1)}{a_{\mathrm{d}f_0}^{2\mathrm{d}f_0}\cdot a_{\mathrm{d}f_1}^{2\mathrm{d}f_1}} \cdot \mathrm{ndisc}(f_0)\cdot \mathrm{ndisc}(f_1)$ }$ ${}\qquad$ (multiplicative.up.to.a.nonpolynomial.factor).

So from this point of view, the 'normalized discriminant' suddenly looks less normalized than the OP's discriminant. 
The only mathematical difference between (multiplicative.up.to.a.polynomial.factor) and (multiplicative.up.to.a.nonpolynomial.factor) I can think of is: 


*

*if $a_{\mathrm{d}f_0}^{2\mathrm{d}f_0}\cdot a_{\mathrm{d}f_1}^{2\mathrm{d}f_1}$ does not divide resultant($f_0,f_1$), where 'divides' refers to the ring generated by the coefficients, then the 'multiplicative constant' in (multiplicative.up.to.a.nonpolynomial.factor) is not a polynomial in the coefficients, while in (multiplicative.up.to.a.polynomial.factor) it is. 


This perhaps might make it possible to characterize the factor the OP is asking about by an algebraic property.
1  Which one could argue is the 'simplest' $\mathrm{auxiliarypolynomial}(f)$ that can be used to 'encode' the logical statement 'there exists a splitting field of $f$ in which there exists a multiple root of $f$ ' in terms of *an equation over the signature of field theory', in the form '$\mathrm{auxiliarypolynomial}(f)=0$'. Though of course 'simplicity-criteria' like 'smallest number of steps to write down the auxiliar polynomial' tend to have little mathematical meaning, and the discriminant is an easy example of this. Slightly more complicated expressions are often simpler to work with. 
A: As Robert said, if you want everything to work in $\mathbb Z[f_0,\ldots,f_m]$, you need that factor. I'll also mention that your polynomial indexing is messed up, you probably meant the sum to go from $j=0$ to $j=m-1$, not $j=1$ to $j=m$.
In any case, things become clearer if you study the theory of resultants and ask when two polynomials have a common root. If they're not monic, you really want to view them as homogeneous polynomials that may have a common root "at infinity" if their leading coefficients both vanish. The discriminant of $f(x)$ is, essentially, the resultant of $f(x)$ and $f'(x)$.
A: The factor $f_0^{2m-2}$ makes the discriminant a polynomial in the coefficients $f_0, \ldots, f_m$.
