Question

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

Answer 1

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

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

Explanation:

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)`