<s>[INST] 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?
 [/INST] $$\frac{5(x-1)(x+2)}{(x-2)(x+3)} = P + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$  gives
$$5(x-1)(x+2)=P(x-2)(x+3)+Q(x+3)+R(x-2)........(1)$$
Now put $x=2, x=-3,x=0$ respectively on both sides of $(1)$ to get,  $Q=4,R=-4,P=5$ respectively. Hence you can reach the desired result.
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