Partial Fractions Decomposition $\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$ explanation repeated factors I am trying to solve the fraction $$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$$
into partial fractions. 
Now, I thought it could be solved into the following
$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{(x-3)^2}$$
but this is apparently incorrect. 
According to the text, the decomposition is 
$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$$
I discussed this with my friend that the fraction first decomposes into 
$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{Bx +C}{(x-3)^2}$$
but I can't see how he derived this. 
I don't understand how he is correct. 

Both decompositions used by your friend and the text are analogous since
$$\frac{Bx+C}{(x-3)^2}=\frac{Bx}{(x-3)^2}+\frac{C}{(x-3)^2}=\frac{B(x-3)+3B}{(x-3)^2}+\frac{C}{(x-3)^2}$$
$$=\frac{B}{x-3}+\frac{3B+C}{(x-3)^2}=\frac{B'}{x-3}+\frac{C'}{(x-3)^2}$$
