Precalculus, Chapter 7, 7.4, Section 7.4, Problem 48

(16x^4)/(2x-1)^3
Since the rational expression is an improper expression, we have to express the fraction as a sum of simpler fractions with the degree of the polynomial in the numerator less than the degree of the polynomial in the denominator.
Let's expand the denominator of the rational expression,
(16x^4)/(2x-1)^3=(16x^4)/(8x^3-12x^2+6x-1)
Dividing using the long division method yields,
(16x^4)/(8x^3-12x^2+6x-1)=2x+3+(24x^2-16x+3)/(8x^3-12x^2+6x-1)
Since the polynomials do not completely divide, so we have to continue with the partial fractions of the remainder expression,
Let, (24x^2-16x+3)/(8x^3-12x^2+6x-1)=A/(2x-1)+B/(2x-1)^2+C/(2x-1)^3
=(A(2x-1)^2+B(2x-1)+C)/(2x-1)^3
=(A(4x^2-4x+1)+2Bx-B+C)/(2x-1)^3
=(x^2(4A)+x(-4A+2B)+A-B+C)/(2x-1)^3
:.(24x^2-16x+3)=x^2(4A)+x(-4A+2B)+A-B+C
equating the coefficients of the like terms,
4A=24 ----- equation 1
-4A+2B=-16 ------ equation 2
A-B+C=3 ------ equation 3
From equation 1,
A=24/4
A=6
Plug the value of A in equation 2,
-4(6)+2B=-16
2B=-16+24
2B=8
B=4
Plug the values of A and B in equation 3,
6-4+C=3
C=3-2
C=1
(24x^2-16x+3)/(8x^3-12x^2+6x-1)=6/(2x-1)+4/(2x-1)^2+1/(2x-1)^3
:.(16x^4)/(2x-1)^3=2x+3+6/(2x-1)+4/(2x-1)^2+1/(2x-1)^3