How do you write #x^4/(x-1)^3# as a partial fraction decomposition?

1 Answer
Oct 27, 2016

The result is
#x^4/(x-1)^3=x+3+6/(x-1)+4/(x-1)^2+1/(x-1)^3#

Explanation:

Since the degree of the numerator is greater than the degree of the denominator, we perform a long division

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

#x^4##color(white)(aaaaaaaaaaaaaaaaaaa)##∣##x^3-3x^2+3x-1#

#x^4-3x^3+3x^2-x # #color(white)(aaaaa)##∣##x+3#

#0+3x^3-3x^2+x#

#color(white)(aaa)##3x^3-9x^2+9x-3#

#color(white)(aaaaa)##0+6x^2-8x+3#

So we get

#x^4/(x-1)^3=x+3+(6x^2-8x+3)/(x-1)^3#

now we form the partial fraction decomposition

#(6x^2-8x+3)/(x-1)^3=A/(x-1)+B/(x-1)^2+C/(x-1)^3#

#(6x^2-8x+3)/(x-1)^3=(A(x-1)^2+B(x-1)+C)/(x-1)^3#

So #6x^2-8x+3=A(x-1)^2+B(x-1)+C#

Let #x=1# then #1=C#

Compare the coefficients of #x^2#

#6=A#

Let #x=0#, then #3=A-B+C#

#B=6+1-3=4#

So the final result is

#x^4/(x-1)^3=x+3+6/(x-1)+4/(x-1)^2+1/(x-1)^3#