Precalculus, Chapter 7, 7.4, Section 7.4, Problem 50

(2x^4+8x^3+7x^2-7x-12)/(x^3+4x^2+4x)
Since the above expression is an improper rational expression , the first step is to divide and express it as a sum of simpler fractions such that the degree of polynomial in the numerator is less than the degree of polynomial in the denominator.
Dividing the above expression using long division method yields,
(2x^4+8x^3+7x^2-7x-12)/(x^3+4x^2+4x)=2x+(-x^2-7x-12)/(x^3+4x^2+4x)
Since the polynomials do not completely divide , we have to continue with the partial fractions of the remainder expression.
Now let's factorize the denominator of the remainder expression,
x^3+4x^2+4x=x(x^2+4x+4)
=x(x+2)^2
Let, (-x^2-7x-12)/(x^3+4x^2+4x)=A/x+B/(x+2)+C/(x+2)^2
=(A(x+2)^2+B(x)(x+2)+Cx)/(x(x+2)^2)
=(A(x^2+4x+4)+B(x^2+2x)+Cx)/(x(x+2)^2)
=(x^2(A+B)+x(4A+2B+C)+4A)/(x(x+2)^2)
:.(-x^2-7x-12)=x^2(A+B)+x(4A+2B+C)+4A
equating the coefficients of the like terms,
A+B=-1 ------- equation 1
4A+2B+C=-7 ------- equation 2
4A=-12
A=-12/4
A=-3
Plug the value of A in equation 1,
-3+B=-1
B=-1+3
B=2
Plug the value of A and B in equation 2 ,
4(-3)+2(2)+C=-7
-12+4+C=-7
C=-7+12-4
C=1
(-x^2-7x-12)/(x^3+4x^2+4x)=(-3)/x+2/(x+2)+1/(x+2)^2
:.(2x^4+8x^3+7x^2-7x-12)/(x^3+4x^2+4x)=2x-3/x+2/(x+2)+1/(x+2)^2