Precalculus, Chapter 7, 7.4, Section 7.4, Problem 46

(x^3+2x^2-x+1)/(x^2+3x-4)
Since the rational expression is an improper expression , we have to express the expression 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.
Dividing using the long division yields,
(x^3+2x^2-x+1)/(x^2+3x-4)=x-1+(6x-3)/(x^2+3x-4)
Polynomials do not completely divide , so we have to continue with the partial fractions of the remainder expression,
Let's factorize the denominator of the remainder fraction,
x^2+3x-4=x^2+4x-x-4
=x(x+4)-1(x+4)
=(x-1)(x+4)
Let (6x-3)/(x^2+3x-4)=A/(x-1)+B/(x+4)
=(A(x+4)+B(x-1))/((x-1)(x+4))
=(Ax+4A+Bx-B)/((x-1)(x+4))
:.(6x-3)=Ax+4A+Bx-B
6x-3=x(A+B)+4A-B
equating the coefficients of the like terms,
A+B=6 ----- equation 1
4A-B=-3 ------ equation 2
Now we have to solve the above equations to get the solutions of A and B,
Adding the equation 1 and 2 yields,
A+4A=6+(-3)
5A=3
A=3/5
Plug the value of A in equation 1 ,
3/5+B=6
B=6-3/5
B=27/5
(6x-3)/(x^2+3x-4)=3/(5(x-1))+27/(5(x+4))
:.(x^3+2x^2-x+1)/(x^2+3x-4)=x-1+3/(5(x-1))+27/(5(x+4))