Rouché's Theorem for $p(z)=z^7-5z^3+12$ 
Let $f$ and $g$ be differentiable on a domain $D$ and suppose that $\gamma$ is a simple closed contour whose inside is contained in D.
If $|f(z)-g(z)|<|f(z)|$ for all $z$ on $\gamma$, then $f$ and $g$ have the same number of zeros inside $\gamma$ (counted including their order).

I was reading an example of application of Rouché's Theorem, where Rouché's theorem was used to show that the polynomial $p(z)=z^7-5z^3+12$ has $0$ roots in $\{z:\mathbb{C}:|z|<1\}$.
What was done was:
Let $g(z)=z^7-5z^3+12$ and let $f(z)=12$. Then for $|z|=1$,
$|f(z)-g(z)|=|z^7-5z^3| \\ \le|z|^7+5|z| \\=1+5\\=6<12=|f(z)|$
Hence by Rouché's Theorem $p(z)=z^7-5z^3+12$ has $7$ roots in $\{z:\mathbb{C}:|z|<2\}$.

I was wondering, what is the purpose to do the step $\le|z|^7+5|z|$? Can't I just jump straight from  $|f(z)-g(z)|=|z^7-5z^3| \\=|1-5|\\=4<12=|f(z)|?$
Secondly, $|f(z)-g(z)|=|-z^7+5z^3|$, is there a reason why they used $|f(z)-g(z)|=|z^7-5z^3|$?
Thirdly, what does it mean by "(counted including their order)"? (From the definition above)
 A: With $|z|=1$, $|f(z)-g(z)|=|z^7-5z^3| = |z^4-5|$, however the last quantity is not equal to $5-1$, as you can take any point with $|z|=1$ (eg, take $z=e^{i\frac{\pi}{4}}$, then the value is $6$). The best upper bound is the one given.
There is no difference between the $|-z^7+5z^3|$ and $|z^7-5z^3|$, since it appears under the $|.|$ (ie, |w| = |-w|).
Order means multiplicity. $z^2$ has a zero of multiplicity $2$ at $0$.
A: It seems that you have applied Rouche's theorem for two regions, the first one is inside the unit circle |z|=1, and the second one is inside the circle |z|=2. Your calculations, regardless of a minor mistake, show that all roots of p(z) lies between these two circles. I use a form of Rouche's theorem, found here ( https://en.wikipedia.org/wiki/Rouche's_theorem ), to provide a similar but complete answer.
1_ For |z|=1 one writes $p(z)=s(z)+t(z)$, $s(z)=z^7-5z^3$, t(z)=12 and then applies triangle inequality on the counter |z|=1 to see that $|s(z)|<1+5<12=|t(z)|$. So p has no roots inside the unit circle (notice that t(z) is constant and has no roots).
2_ Mimic the method of step 1, this time for $p(z)=u(z)+v(z)$, $u(z)=-5z^3+12$, $v(z)=z^7$, |z|=2 and see that $|u(z)| < 5*8+12=52 < 128 = |v(z)|$. Thus the total roots of p(z) must lie inside the circle |z|=2.
Now it is evident from steps 1 and 2 that all roots of p(z) are between the circles |z|=1 and |z|=2.
