How do you write the partial fraction decomposition of the rational expression (2x^3-x^2+x+5)/(x^2+3x+2)?

1 Answer
Aug 7, 2016

(2x^3-x^2+x+5)/(x^2+3x+2) = 2x-7+1/(x+1)+17/(x+2)

Explanation:

(2x^3-x^2+x+5)/(x^2+3x+2)

=(color(blue)(2x^3+6x^2+4x)color(green)(-7x^2-21x-14)+18x+19)/(x^2+3x+2)

=(color(blue)(2x(x^2+3x+2))color(green)(-7(x^2+3x+2))+18x+19)/(x^2+3x+2)

=2x-7+(18x+19)/(x^2+3x+2)

Focusing on the remaining rational expression:

(18x+19)/(x^2+3x+2)

=(18x+19)/((x+1)(x+2))

=A/(x+1)+B/(x+2)

Using Heaviside's cover-up method we find:

A=(18(-1)+19)/((-1)+2) = 1/1 = 1

B=(18(-2)+19)/((-2)+1) = (-17)/(-1) = 17

So:

(18x+19)/(x^2+3x+2)=1/(x+1)+17/(x+2)

Putting it all together:

(2x^3-x^2+x+5)/(x^2+3x+2) = 2x-7+1/(x+1)+17/(x+2)