What's the partial fraction of

[(x-2)(x-3) ] / [(x+4)(x-5)]

lets partial fractions be A/(x+4) and B/(x-5)
therefore,
[(x-2)(x-3) ] / [(x+4)(x-5)] = A/(x+4) + B/(x-5) + C(x+4)(x-5)

multiply by (x+4)(x-5),

(x-2)(x-3) = A(x-5) + B/(x+4) ----------- (1)

substituting x = 5 in (1),

3X2 = 9B
B = 2/3

substituting x = -4 in (1),

-6X-7 = -9A
A = -14/3
by comparing coefficient of constant terms,

C = -3/10

therefore partial fraction are

-14/3(x+4) , 2/3(x-5) and -3/10

It is an improper fraction so no partial fractions.

expand the expression in numerator and denominator.
Then, divide numertr by demtr.