Practical calculation of polynomial expansions in powers of a linear form $x-r$ In The Fundamental Theorem of Algebra: A Visual Approach by
Daniel J. Velleman there appears the comment

For example, near the root at $−1$ in Figure 3, the colors have been rotated $180$ degrees, so that red is to the left of the root rather than to the right. This is because when the polynomial $f(z)$ of Figure 3 is expanded in powers of $z + 1,$ the coefficient of $z + 1$ is a negative real number.)


$-1$ is a root of multiplicity $1$ of the polynomial 
$$z^8-2z^7+2z^6-4z^5+2z^4-2z^3-5z^2+4z-4.$$
Expanding in powers of seems to correspond to a change of variables - parenthetically, is there a name for this so as to avoid misunderstandings with Taylor expansion? - so that
$$(z + 1) \underbrace{(z^7 - 3 z^6 + 5 z^5 - 9 z^4 + 11 z^3 - 13 z^2 + 8 z - 4)}_{g(z)}$$
leads to a reformulation of $g(z)$ as
$$g(z) = a_0 + a_1 (z +1) + a_2(z +1)^2 + \ldots$$
This is probably nothing one should attempt without professional help, but I can't even seem to figure out how to get it done with Wolfram Alpha...

How do you go about doing a change of variables as this one?

In this case I would like to see how "the coefficient of $z+1$ is a negative real number" as a measure of success.
 A: EDITED TO EXPLAIN WHY TAYLOR EXPANSION WORKS
You have actually mentioned in your answer what you need: the expansion in powers of $z+1$ of $f(z)$ is nothing more than Taylor expansion around $-1$. To do it, you just use that
$$f(z)=\sum_{i=0}^8 \frac{f^{(i)}(-1)}{i!}(z+1)^i.$$
Explicitly, we have
$$f(z)=-54(z+1)+153(z+1)^2-216(z+1)^3+192(z+1)^4-114(z+1)^5+44(z+1)^6-10(z+1)^7+(z+1)^8.$$
You can expand and collect to check that this expression is indeed $f(z)$ (in Matlab you can write collect(polynomial,z))

ADDED:
Suppose $f(X)=a_nX^n+\ldots +a_1X+a_0$ and you want to write the same polynomial but centered at the number $c$, i.e., as
$$f(X)=b_n(X-c)^n+\ldots+b_1(X-c)+b_0.$$
So, what we want is to find the coefficients $b_i$ from our initial expression for $f$. Observe that $b_0$ is just $f(c)$. This evaluation idea is very convenient; can we apply it to find the other $b_i$?
Well, if we look at $f(X)-b_0$ we see we have a multiple of $X-c$, 
$$f(X)-b_0=(b_n(X-c)^{n-1}+\ldots+b_1)(X-c),$$
and if we divide by $X-c$ we get
$$f_1(X):=\frac{f(X)-b_0}{X-c}=b_n(X-c)^{n-1}+\ldots+b_2(X-c)+b_1,$$
so now we can evaluate $f_1(c)$ to get $b_1$.
As you see, this process of getting $b_i$ from $f_i(X)$ can be continued until we have all the desired coefficients.
But let us see, how does the procedure work? We are getting rid of a constant, and then we are reducing the degree of every $(X-c)^i$ by one. Doesn't this sound like taking the derivative?
If instead of substracting and dividing we just derive, we get instead
$$f^{(1)}(X)=nb_n(X-c)^{n-1}+\ldots+2b_2(X-c)+b_1,$$
$$f^{(2)}(X)=n(n-1)b_n(X-c)^{n-2}+\ldots+3\cdot2(X-c)+2b_2,$$
and so on. We see that an artificial factor of $i!$ appears when we evaluate $f^{(i)}(c)=i!b_i$ to get $b_i$, so we need to divide by it to get 
$$b_i=\frac{f^{(i)}(c)}{i!}.$$
Therefore we have shown (and understood)
$$f(z)=\sum_{i=0}^n \frac{f^{(i)}(c)}{i!}(X-c)^i.$$
A: Let $\,z+1=w\,$ so that $\,z=w-1\,$, then:
$$
 z^7 - 3 z^6 + 5 z^5 - 9 z^4 + 11 z^3 - 13 z^2 + 8 z - 4 \\[5px]
 = (w-1)^7 - 3(w-1)^6 + 5 (w-1)^5-9(w-1)^4+11(w-1)^3-13(w-1)^2+8(w-1) - 4 \\[5px]
 = w^7 - 10 w^6 + 44 w^5 - 114 w^4 + 192 w^3 - 216 w^2 + 153 w - 54 \\[5px]
 = (z+1)^7 - 10 (z+1)^6 + 44 (z+1)^5 - 114 (z+1)^4 + 192 (z+1)^3 - 216 (z+1)^2 + 153 (z+1) - 54
$$
The  calculation can be verified in Wolfram Alpha, as shown under "expanded form".
