How do you write the partial fraction decomposition of the rational expression #(4x^2+3)/((x-5)^3)#?

1 Answer
Jul 31, 2018

#(4x^2+3)/(x-5)^3=4/(x-5)+40/(x-5)^2+103/(x-5)^3#

Explanation:

Let ,

#(4x^2+3)/(x-5)^3=A/(x-5)+B/(x-5)^2+C/(x-5)^3 ...tocolor(red)((M)#

#:.4x^2+3=A(x-5)^2+B(x-5)+C#

#:.4x^2+3=A(x^2-10x+25)+B(x-5)+C#

#:.4x^2+3=Ax^2-10Ax+25A+Bx-5B+C#

#:.4x^2+0*x+3=Ax^2+x(B-10A)+(25A-5B+C)#

Comparing coefficients of #x^2 ,x and# free term,we get

#color(blue)(A=4......................to(1)#

#B-10A=0.........,.to(2)#

#25A-5B+C=3..to(3)#

Substitute #color(blue)(A=4# into #(2)#

#:.B-10(4)=0=>color(blue)(B=40#

Again subst. #A=4 and B=40# into #(3)#

#:.25(4)-5(40)+C=3#

#:.C=3-100+200=>color(blue)(C=103#

Hence, from #color(red)((M))#,we have

#(4x^2+3)/(x-5)^3=color(blue)(4)/(x-5)+color(blue)(40)/(x-5)^2+color(blue)(103)/(x-5)^3#