How do you write the partial fraction decomposition of the rational expression #(X^2-5x+6)/(X^3-X^2+2X)#?

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Answer 1 · 2015-12-30T04:28:43

#3/x-(2(x+1))/(x^2-x+2)#

Factor out an #x# in the denominator

#(x^2-5x+6)/(x(x^2-x+2))=A/x+(Bx+C)/(x^2-x+2)#

Multiply both sides by #x(x^2-x+2)#

#x^2-5x+6=A(x^2-x+2)+x(Bx+C)#

Distribute on right hand side

#x^2-5x+6=Ax^2-Ax+2A+Bx^2+Cx#

Factor right hand side

#x^2-5x+6=x^2(A+B)+x(-A+C)+2A#

Equate coefficients on the left hand side with the right

#2A=6#

#-A+C=-5#

#A+B=1#

Solving we get

#A=3#

#B=-2#

#C=-2#

Putting it all together

#(x^2-5x+6)/(x^3-x^2+2x)=3/x+(-2x-2)/(x^2-x+2)#

Pull out negative in #-2x-2#

#(x^2-5x+6)/(x^3-x^2+2x)=3/x-(2x+2)/(x^2-x+2)#

Factor out a #2# in #2x+2#

#(x^2-5x+6)/(x^3-x^2+2x)=3/x-(2(x+1))/(x^2-x+2)#