To write the given expression into partial fractions we think about factorizing the denominator.
Let us factorize the denominator
#color(blue)(x^3-2x^2-x+2)#
#=color(blue)(x^2(x-2)-(x-2))#
#=color(blue)((x-2)(x^2-1))#
Applying the identity of polynomials:
#color(orange)(a^2-b^2=(a-b)(a+b))#
we have:
#color(blue)(x^3-2x^2-x+2)#
#=color(blue)((x-2)(x^2-1^2))#
#=color(blue)((x-2)(x-1)(x+1))#
Let us decompose the rational expression by finding #A,B,and C#
#color(brown)(A/(x-2)+B/(x-1)+C/(x+1))=color(green)((3x)/(x^3-2x^2-x+2))#
#color(brown)(A/(x-2)+B/(x-1)+C/(x+1))#
#=color(brown)((A(x-1)(x+1))/(x-2)+(B(x-2)(x+1))/(x-1)+(C(x-2)(x-1))/(x+1))#
#=(A(x^2-1))/(x-2)+(B(x^2+x-2x-2))/(x-1)+(C(x^2-x-2x+2))/(x+1)#
#=(A(x^2-1))/(x-2)+(B(x^2-x-2))/(x-1)+(C(x^2-3x+2))/(x+1)#
#=(Ax^2-A+Bx^2-Bx-2B+Cx^2-3Cx+2C)/((x-2)(x-1)(x+1)#
#=color(brown)(((A+B+C)x^2+(-B-3C)x+(-A-2B+2C))/((x-2)(x-1)(x+1))#
#=color(brown)(((A+B+C)x^2+(-B-3C)x+(-A-2B+2C))/((x-2)(x-1)(x+1))=color(green)((3x)/(x^3-2x^2-x+2))#
Then ,
#rArrcolor(brown)((A+B+C)x^2+(-B-3C)x+(-A-2B+2C))=color(green)(3x)#
We have a system of three equations with three unknowns #A,B and C#
#A+B+C=0# eq1
#-B-3C=3# eq2
#-A-2B+2C=0# eq3
Starting to solve the system
eq2:#-B-3C=3rArr-B=3+3CrArrcolor(red)(B=-3-3C)#
Substituting #B# in eq1 we have:
#A+B+C=0#
#A-3-3C+C=0rArrA-3-2C=0rArrcolor(red)(A=3+2C)#
Substituting #B and C#in eq3 we have:
#-A-2B+2C=0# eq3
#rArr-(color(red)(3+2C))-2(color(red)(-3-3C))+2C=0#
#rArr-3-2C+6+6C+2C=0#
#rArr+3+6C=0#
#rArr6C=-3#
#rArrcolor(red)(C=-1/2)#
#color(red)(B=-3-3C)=-3-3color(red)(-1/2)=-3+3/2#
#color(red)(B=-3/2#
#color(red)(A=3+2C)=3+2(-1/2)=3-1#
#color(red)(A=2)#
Let us substitute the values:
#color(green)((3x)/(x^3-2x^2-x+2))=color(brown)(color(red)2/(x-2)+(color(red)(-3/2))/(x-1)+color(red)((-1/2))/(x+1))#
Therefore,
#(3x)/(x^3-2x^2-x+2)=2/(x-2)-3/(2(x-1))-1/(2(x+1))#
