How do you write the fractional decomposition of #(5x^2 + 7x + 3)/(x^3 + x)#?

1 Answer
Feb 16, 2017

#(5x^2+7x+3)/(x^3+x) = color(green)(3/x+(1+7/2i)/(x+i)+(1-7/2i)/(x-i))#

Explanation:

First we need to find the linear factors of the denominator (note that 2 of these factors only exist as complex values):
#x^3+x=x(x^2+1)=color(blue)(x(x+i)(x-i))#

We need to find values #A, B, and C# such that
#(5x^2+7x+3)/(x^3+x)=A/x +B/(x+i)+C/(x-i)#

#color(white)("XXXXX")=(A(x+i)(x-i)+B(x)(x-i)+C(x)(x+i))/(x(x+i)(x-i))#

#color(white)("XXXXX")=(Ax^2+A+Bx^2-Bix+Cx^2-Cix)/(x^3+x)#

#rarr5x^2+7x+3=((A+B+C)x^2+(-Bi+Ci)x+A#
#rarrcolor(white)("XXXXX"){([1]color(white)("X")A+B+C=5),([2]color(white)("X")-Bi+Ci=7),([3]color(white)("X")A=3):}#

Combining [1] and [3] we have
#color(white)("XXXXX")[4]color(white)("X")B+C=2#

and from [2] we can
#color(white)("XXXXX")[5]color(white)("X")B-C=7i#

Adding [4] and [5] then dividing by 2:
#color(white)("XXXXX")[6]color(white)("X")B=1+7/2i#

Then substituting back into [4]
#color(white)("XXXXX")C=1-7/2i#