so I have
x^2 / (x-1)(x-2)
The answer is 1 - 1/(x-1) + 4/(x-2).
Where did the 1 come from. I did the whole solving for A and B thing and got the same answer without the 1.
Thanks.
$\displaystyle \frac {x^2}{(x-1)(x-2)}=\frac {x^2}{x^2-3x+2}= 1+\frac{3x-2}{x^2-3x+2}=1+\frac{3x-2}{(x-1)(x-2)}$
now resolve$\displaystyle \frac{3x-2}{(x-1)(x-2)}$ into partial fraction
$\displaystyle \frac{3x-2}{(x-1)(x-2)}=\frac{- 1}{(x-1)} + \frac{4}{(x-2)}$
$\displaystyle therefore\quad \frac {x^2}{(x-1)(x-2)}=1-\frac{ 1}{(x-1)} + \frac{4}{(x-2)}$