partial fraction (with factoring involved) in recurrence relation problem i want to factor the denominator of the function $f(x)=4-23x+21x^2/(1-2x-3x^2)(1-3x)$ so that it takes the form $(1-ax)(1-bx)(1-3x)$. 
What I've got is $4-23x+21x^2/(1-3x)(1+x)(1-3x)$ but the denominator has $(1-ax)(1+bx)$ form here.I've been thinking this for an hour now but the question doesn't really make sense to me .I can't think of any other possible denominator form other than $(1-3x)(1+x)(1-3x)$. One person suggests me $(1-3x)(1-[-1]x)(1-3x)$ but Im not sure if this is correct because I need this function to decompose partial fractions but then the problem becomes somewhat complicated (due to algebra involved)
The question further asks me to b)decompose partial fractions (the constants should be all small integers) and c)to extract the coefficient $[x^n]f(x)$ using the result above.
Anyways, using the my equation $4-23x+21x^2/(1-3x)(1+x)(1-3x)$, I tried to decompose fraction, and $A/(1+x) + B/(1-3x) + C/(1-3x)^2$ and $A(1-3x)(1-3x)+B(1-3x)(1+x)+C(1+x)$ and by matching the coefficients $21=9A-3B$, $-23=-6A-2B+C, 4=A+B+C$
I just need to solve for coefficients $A, B, C$ but when I solve for A in $21=9A-3B$, the A seems to cancel out each other and I get 21=0 which is definitely wrong.
Can anyone please clarify this? Any help would be very much appreciated Thank you

The way of answering the question is definitely $b=-1$, unless you were told to find the coefficients in $\mathbb{R_+}$, which would make the question wrong.
Now concerning the decomposition, you must have made a mistake in calculating the coefficients. What you should get is:
$$ f(x) = \frac{4-23x+21x^2}{(1-3x)(1+x)(1-3x)} = \frac{3}{1+x}+ \frac{2}{1-3x}-\frac{1}{(1-3x)^2} $$
Which does indeed give small integers for $A, B, C$.
