Trouble with factorising a polynomial I'm supposed to show that:  
$$y=\frac{5(x-1)(x+2)}{(x-2)(x+3)} = P + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$  
and the required answers are: $$ P=5, Q=4, R=-4 $$  
I tried to solve this with partial fractions like so:  
$$5(x-1)(x+2) = A(x+3) + B(x-2)$$  
$\implies$ $A$=4, $B$=-4
$\implies$ $Q$=4, R=-4  
But where does $P$=5 come from?
Or should I have first multiplied out the numerator and denominator and then used long division to solve?

You are given the wrong expression. After some calculation I figure out that in order for these two expressions to be equivalent,
$$\frac{5(x-1)(x-2)}{(x-2)(x+3)} =  5 + \frac{4}{(x-2)} + \frac{-4}{(x+3)}$$ 
instead of having $\frac{5(x-1)(x-2)}{(x-2)(x+3)}$, it should be
$$\frac{5(x-1)(x+2)}{(x-2)(x+3)}$$
Now solve it again it would work well.
My approach to solving it is 
$$\frac{5(x-1)(x+2)}{(x-2)(x+3)} = \frac{5x^2+5-10}{(x-2)(x+3)} =\frac{\frac{5x^2+5-10}{(x-2)}}{(x+3)}$$
$$\frac{\frac{5x^2+5-10}{(x-2)}}{(x+3)} = \frac{5x+15+\frac{40}{(x-2)}}{(x+3)}$$
$$\frac{5x+15+\frac{40}{(x-2)}}{(x+3)} = 5+\frac{40}{(x-2)(x+3)}$$
$$5+\frac{40}{(x-2)(x+3)}=5 + \frac{4}{(x-2)} + \frac{-4}{(x+3)}$$
