How do you write the partial fraction decomposition of the rational expression $(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))$ ?

Question

1506 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-01T04:44:02

$(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))=2+1/(x-3)-3/(x+2)+2/(x+2)^2$

Before we express in partial fractions, ensure that the degree of numerator is less than that of denominator.

$(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))$

= $(2x^3+2x^2-9x-20)/(x^3+x^2-8x-12)$

= $2+(7x+4)/(x^3+x^2-8x-12)$

= $2+(7x+4)/((x-3)(x+2)(x+2))$

= $2+(7x+4)/((x-3)(x+2)^2)$

Now let $(7x+4)/((x-3)(x+2)^2)-=A/(x-3)+B/(x+2)+C/(x+2)^2$

or $(7x+4)/((x-3)(x+2)^2)=(A(x+2)^2+B(x-3)(x+2)+C(x-3))/((x-3)(x+2)^2)$

if $x=3$ , then $25A=25$ i.e. $A=1$

if $x=-2$ , then $-5C=-10$ i.e. $C=2$

and comparing coefficients of $x^2$ on both sides

$A+B+C=0$ i.e. $B=-3$

Hence

$(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))=2+1/(x-3)-3/(x+2)+2/(x+2)^2$

How do you write the partial fraction decomposition of the rational expression (2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))?