question about partial fractions why $\frac{6x^2+19x+15}{(x-1)(x-2)^2}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-2)^2}$ can anyone tell me why
$$\frac{6x^2+19x+15}{(x-1)(x-2)^2}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$$
I don't undestand the $\frac{C}{(x-2)^2}$ and also what is wrong according to basic math rules if I write the equation in the following way
$$\frac{6x^2+19x+15}{(x-1)(x-2)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-2}$$
Thanks

Two things wrong with what you propose. 


*

*Those last two terms could be combined into one, $${B\over x-2}+{C\over x-2}={D\over x-2}$$ 

*It just doesn't work ---- what would you get for $B$ and $C$ if you tried to do $${1\over(x-2)^2}={B\over x-2}+{C\over x-2}$$
