Because a rational function in the form:
#A/(x-4)+B/(x+5)#
has a numerator of first degree in #x# as we can see by developing the expression:
#A/(x-4)+B/(x+5) = (A(x+5)+B(x-4))/((x-4)(x+5)) = ( (A+B)x + (5A-4B))/((x-4)(x+5))#
whereas the original function:
#x^2/((x-4)(x+5))#
has a numerator of second degree in #x#.
In fact we can see that for any #A# and #B# we have:
#lim_(x->oo) A/(x-4)+B/(x+5) = 0#
while:
#lim_(x->oo) x^2/((x-4)(x+5)) = 1#
To perform partial fraction decomposition, the numerator must have a degree in #x# lower that the denominator, and the correct way to proceed is:
#x^2/((x-4)(x+5)) = x^2/(x^2+x-20) = (x^2+x-20-x+20)/(x^2+x-20) =1 - (x-20)/((x-4)(x+5))#
We can now pose:
#(x-20)/((x-4)(x+5)) = A/(x-4)+B/(x+5)#
#(x-20)/((x-4)(x+5)) = ( (A+B)x + (5A-4B))/((x-4)(x+5))#
#{(A+B=1),(5A-4B=20):}#
#{(A=8/3),(B=-5/3):}#
and conclude that:
#x^2/((x-4)(x+5)) = 1-8/(3(x-4))+5/(3(x+5))#