How do you write the partial fraction decomposition of the rational expression #(10x + 2)/ (x^3 - 5x^2 + x - 5)#?

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Answer 1 · 2016-01-18T23:01:56

#(10x+2)/(x^3-5x^2+x-5) = (-2x)/(x^2+1)+2/(x-5)#

#=-1/(x-i)-1/(x+i)+2/(x-5)#

First factor the denominator by grouping:

#x^3-5x^2+x-5#

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

#=x^2(x-5)+1(x-5)#

#=(x^2+1)(x-5)#

If we stick with Real coefficients for now, then these factors will be the denominators of the partial fraction decomposition.

So we want to solve:

#(10x+2)/(x^3-5x^2+x-5) = (Ax+B)/(x^2+1)+C/(x-5)#

#=((Ax+B)(x-5)+C(x^2+1))/(x^3-5x^2+x-5)#

#=((A+C)x^2+Bx+(C-5B))/(x^3-5x^2+x-5)#

Equating coefficients of #x^2#, #x# and the constant term we find:

(i) #A+C = 0#

(ii) #B-5A=10#

(iii) #C-5B=2#

Combining these using the recipe (i) - (iii) - 5(ii) we find:

#A+C-C+5B-5B+25A = 0-2-50 = -52#

So: #A = -52/26 = -2#

Hence: #C = 2#

and #B=10+5A = 10-10=0#

So:

#(10x+2)/(x^3-5x^2+x-5) = (-2x)/(x^2+1)+2/(x-5)#

Next, if we allow Complex coefficients, then we can factor #(x^2+1)# as #(x-i)(x+i)#, hence:

#(-2x)/(x^2+1) = -1/(x-i) - 1/(x+i)#