The #Dr.=8x^2-6x+1=(2x-1)(4x-1)#
Both the factors are linear polys. Hence, the reqd. decomposition
is given by,
#x/(8x^2-6x+1)=A/(2x-1)+B/(4x-1), where, A,B in RR#.
To determine #A,&,B#, Heavyside's Cover-up Method is useful :-
#A=[x/(4x-1)]_(x=1/2) =1/2/(4*1/2-1)=1/2#.
#B=[x/(2x-1)]_(x=1/4)=1/4/(2*1/4-1)=1/4/(-1/2)=-1/2#.
Therefore,
#x/(8x^2-6x+1)=(1/2)/(2x-1)-(1/2)/(4x-1)#.
Method II :-
When there are only two linear factors in #Dr.#, this decomposition can easily be derived as under :
Observe that, #(4x-1)-(2x-1)=2x#
#rArr Nr. =x=1/2{(4x-1)-(2x-1)}#. Therefore,
#x/(8x^2-6x+1)=1/2[{(4x-1)-(2x-1)}/((4x-1)(2x-1))]#
#=1/2[cancel(4x-1)/(cancel(4x-1)(2x-1))-cancel(2x-1)/((4x-1)cancel(2x-1))]#
#=1/2[1/(2x-1)-1/(4x-1)]#, as obtained earlier!
Enjoy Maths.!
