Precalculus, Chapter 7, 7.4, Section 7.4, Problem 13

(4x^2+3)/(x-5)^3
To decompose this fraction, consider the factor in the denominator. The factor is repeated three times. So the partial fraction of this will have denominator in increasing exponent of the factor.
A/(x-5) , B/(x-5)^2 , and C/(x-5)^3
Add these three fractions and set it equal to the given fraction.
(4x^2+3)/(x-5)^3=A/(x-5) + B/(x-5)^2+C/(x-5)^3
To solve for the values of A, B and C, eliminate the fractions in the equation. So, multiply both sides by the LCD.
(x-5)^3*(4x^2+3)/(x-5)^3= (A/(x-5)+B/(x-5)^2+C/(x-5)^3)*(x-5)^3
4x^2+3=A(x-5)^2+B(x-5)+C
4x^2+3=A(x^2-10x+25)+B(x-5)+C
Then, group together the terms with x^2, the terms with x and the constants.
4x^2+3=Ax^2 - 10Ax + 25A +Bx - 5B+C
4x^2+3=Ax^2 + (-10A + B)x + (25A-5B+C)
Set the coefficient of x^2 at the left side equal to the coefficient of x^2 at the right side.
4=A This is the value of A.
Then, set the coefficient of x at the left side equal to the coefficient of x at the right side.
0=-10A+B (Let this be EQ1.)
And, set the constant at the left side equal to the constant at the right side.
3=25A - 5B + C (Let this be EQ2.)
To get the value of B, plug-in A=4 to EQ1.
0=-10A+B
0=-10(4) + B
0=-40+B
40=B
And to get the value of C, plug-in A=4 and B=40 to EQ2.
3=25A - 5B + C
3=25(4)-5(40)+C
3=100-200+C
3=-100+C
3+100=C
103=C
So the partial fraction decomposition of the given rational expression is:
4/(x-5) + 40/(x-5)^2 + 103/(x-5)^3

To check, express the fractions with same denominators.
4/(x-5)+40/(x-5)^2+103/(x-5)^3
=4/(x-5)*(x-5)^2/(x-5)^2 + 40/(x-5)^2*(x-5)/(x-5)+103/(x-5)^3
=(4(x^2-10x+25))/(x-5)^3 + (40(x-5))/(x-5)^3)+103/(x-5)^3
Now that they have same denominators, proceed to add them.
= (4(x^2-10x+25) + 40(x-5) + 103)/(x-5)^3
=(4x^2-40x+100+40x-200+103)/(x-5)^3
=(4x^2+3)/(x-5)^3

Therefore, (4x^2+3)/(x-5)^3=4/(x-5)+40/(x-5)^2+103/(x-5)^3 .