Precalculus, Chapter 7, 7.4, Section 7.4, Problem 45

(2x^3-x^2+x+5)/(x^2+3x+2)
Since the rational expression is an improper rational expression, so we have to express it as sum of simpler fractions with degree of polynomial in the numerator less than the degree of polynomial in the denominator.
Dividing using long division yields,
(2x^3-x^2+x+5)/(x^2+3x+2)=2x-7+(18x+19)/(x^2+3x+2)
Since the polynomial don not completely, so we have to continue with the partial fractions of the remainder expression,
Let's factorize the denominator of the remainder expression,
x^2+3x+2=x^2+x+2x+2
=x(x+1)+2(x+1)
=(x+1)(x+2)
Let(18x+19)/(x^2+3x+2)=A/(x+1)+B/(x+2)
=(A(x+2)+B(x+1))/((x+1)(x+2))
=(Ax+2A+Bx+B)/((x+1)(x+2))
=(x(A+B)+2A+B)/((x+1)(x+2))
:.(18x+19)=x(A+B)+2A+B
Equating the coefficients of the like terms,
A+B=18 --------- equation 1
2A+B=19 -------- equation 2
Now solve the above equations to get the solutions of A and B,
Subtract equation 1 from equation 2
2A-A=19-18
A=1
plug the value of A in equation 1,
1+B=18
B=18-1
B=17
(18x+19)/(x^2+3x+2)=1/(x+1)+17/(x+2)
:.(2x^3-x^2+x+5)/(x^2+3x+2)=2x-7+1/(x+1)+17/(x+2)