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 Cin 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))
