Easy ways to find partial fraction representation? (via a concrete example) In a homework assignment (about generating functions) the students find themselves having to expand $\frac{3-7x+9x^{2}-3x^{3}}{\left(1-x\right)^{4}}$ intro partial fractions. Using some automated tool (e.g. Wolfram alpha) the expansion is found right away: $\frac{3}{1-x}-\frac{2}{\left(1-x\right)^{3}}+\frac{2}{\left(1-x\right)^{4}}$.
However, sending the students to use Wolfram alpha instead of computing by themselves seems problematic to me. However, I have no idea how the partial fraction representation can be found without a lot of dirty work involved, which will ensure the students hate the exercise instead of learning from it.
My question - is there some "nice" way of doing such things I am simply unaware of? Of course, the best answer is a demonstration on the concrete case of $\frac{3-7x+9x^{2}-3x^{3}}{\left(1-x\right)^{4}}$ - but only give it if your method is simple enough so that writing it down is not a lot of work...
 A: In this particular case, I think the quickest way is to introduce a new variable $t=1-x$. So just substitute $x=1-t$ and expand:
$$
\frac{3-7x+9x^{2}-3x^{3}}{\left(1-x\right)^{4}}=
\frac{3-7(1-t)+9(1-t)^{2}-3(1-t)^{3}}{t^{4}}=
\frac{2-2t+0t^2+3t^3}{t^4}=
\frac{2}{t^4} - \frac{2}{t^3} + \frac{3}{t}.
$$
Now put $t=1-x$ back, and you have the partial fractions decomposition.
A: $${3-7x+9x^2-3x^3\over(1-x)^4}={A\over1-x}+{B\over(1-x)^2}+{C\over(1-x)^3}+{D\over(1-x)^4}$$ 
$$3-7x+9x^2-3x^3=A(1-x)^3+B(1-x)^2+C(1-x)+D$$ 
Substitute $x=1$ to get $D$. Differentiate and substitute $x=1$ to get $C$. Differentiate again and let $x=1$ to get $B$. Guess what? Differentiate again and put $x=1$ to get $A$. 
A: When you do that thing, you use the fact that the polynomials $1$,$(1-x)$, $(1-x)^2$, $(1-x)^3$ and $(1-x)^4$ form a basis of the vector space of polynomials of degree $\le 4$, and thus you try to express $3-7x+9x^2-3x^3$ as a linear combination of $(1-x)$, ... $(1-x)^4$. Clearly since the polynomial you're dealing with has degree $3$, so that you don't need the degree $4$ term (i.e. we are sure that the coefficient in front of it is $0$, so we don't need a term of the form "$E(1-x)^4$"). This leaves you with the following system :
$$
3-7x+9x^2-3x^3 = A + B(1-x) + C(1-x)^2 + D(1-x)^3
$$
The "dirty work" you are saying is always going to be linear algebra, and unless you do calculation by hand (and this might involve tricks, like plugging in numbers, differentiating, manipulations, etc), the only easy trick (the one in which you just "apply a formula") there is is to form a matrix and invert it. To see how this could go, write your polynomial in the form $(3,-7,9,-3)$, which is its writing over the standard basis $\{1,x,x^2,x^3\}$. With that in mind, expand the polynomials on the RHS to find
$$
A + B(1-x) + C(1-x)^2 + D(1-x)^3 = (A+B+C+D) + (-B-2C-3D)x + (C+D)x^2 + (-D)x^3.
$$
This means that a polynomial written over the basis $\{1,(1-x),(1-x)^2,(1-x)^3\}$ with coefficients $(A,B,C,D)$ gets mapped to a polynomial written over the basis $\{1,x,x^2,x^3\}$ with coefficients $(A+B+C+D,-B-2C-3D,C+D,-D)$. This means
$$
\begin{bmatrix} 3 \\ -7 \\ 9 \\ -3 \\ \end{bmatrix} = 
\begin{bmatrix}
A+B+C+D \\
-B-2C-3D \\
C+3D \\
-D \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 & 1 & 1 \\
0 & -1 & -2 & -3 \\
0 & 0 & 1 & 3 \\
0 & 0 & 0 & -1 \\
\end{bmatrix}
\begin{bmatrix}
A \\ B \\ C \\ D \\
\end{bmatrix}
$$
So that if you invert the matrix and multiply it by your vector of coefficients you get the new vector $(A,B,C,D)$ with your requested coefficients. Dividing the first equation above gives you what you wanted.
This is far from being the most quick way to do it by hand since we humans are very lazy to invert matrices =) But as Gerry Myerson and Geoff Robinson have shown, there are ways to find some coefficients quickly and that simplifies the work. 
Hope that helps,
A: One thing that students sometimes find appealing (and I did when I first saw it), is the use of Taylor polynomials. Put $f(x) = (3 - 7x+9x^2 -3x^3)$, and then write
$f(x)$ as $f(1) + f^{\prime}(1)(x-1) + \frac{f^{"}(1)}{2!}(x-1)^{2} + \frac{f^{(3)}(1)}{3!}(x-1)^{3}$. Of course, the same trick will work whenever you have a rational function of the form $\frac{g(x)}{(x-a)^d}$ to deal with. If $g(x)$ has degree $n,$ just write it as $\sum_{j=0}^{n} \frac{g^{(j)}(a)}{j!}(x-a)^{j}$.
