It's the same as you would normally do it, except with a third unknown - typically 'C'. Also your working is wrong, its just A B C, one for each denominator factor. From there you could use your own method of getting the simultaneous equations.
You have three linear factors and the original numerator is of order 3 or less. So you can split it into three separate fractions each with a constant numerator.
You have three linear factors and the original numerator is of order 3 or less. So you can split it into three separate fractions each with a constant numerator.
OK - get rid of the A - then write the right hand side over the common denominator.
B (x + 2)(x + 1) + C (x + 3)(x + 1) + D(x + 3)(x + 2) will be the numerator.
Can you go from here?
I got that bit, and i assume you get a simultaneous equation but i don't understand how you get two separate equations when there's only a 2 on the numerator
I got that bit, and i assume you get a simultaneous equation but i don't understand how you get two separate equations when there's only a 2 on the numerator
Put x = -1 then
2 = D(-1 + 3)(-1 + 2)
Then repeat for other values of x to get values of B and C