# Simple partial fraction expansion

I would like to ask for help solving a simple Partial fraction example: (It contains complex roots)

$\frac{1}{(x^3 - 3x^2 + 4x - 2)}$

According to my book we can factor it as a product of the real roots and quadratic factors:

$\frac{1}{(x-1)(x^2-2x+2)}$

Quadratic form should adopt the following scheme:

$\frac{ax+b}{x^2+px+q}$

Using the same system as in the case that all the roots are real:

$\frac{a}{x-1}+\frac{ax+b}{x^2-2x+2}==>\frac{a(x^2-2x+2)+(ax+b)(x-1)}{(x-1)(x^2-2x+2)}$

And then we solve the equation:

$a(x^2-2x+2)+(ax+b)(x-1) = 1$

The book expose the solution:

$\frac{1}{x-1}+\frac{1-x}{x^2-2x+2}$

I just can't figure out how they do the transition. I am learning to use "Inverse Laplace Partial Fractions with Complex Roots" by converting the quadratic form to a product of complex roots:

$(x^3 - 3x^2 + 4x - 2)=(x-1)(x-1-i)(x-1+i)$

But it is just way out of the current scope (is not explained in the book, I found it during research). There must be some simple method/analysis but I just can't see it.

• As Adi Dani said your two $a$ are not the same. And you need to prove (by induction) that if $\displaystyle f(x) = \frac{Q(x)}{\prod_{i=1}^I (x-a_i) \prod_{j=1}^J (x^2+b_j x+ c_j)}$ then there exists some constants $A_i,B_j,C_j$ and a polynomial $R$ such that $\displaystyle f(x) = R(x) + \sum_{i=1}^I \frac{A_i}{x-a_i}+\sum_{j=1}^J \frac{B_jx+C_j}{x^2+b_j x+ c_j}$ – reuns Dec 10 '16 at 17:25
• Neat function! More clear than in my book, thanks! – Charlie Dec 10 '16 at 17:31

## 1 Answer

You need to start from $$\frac{a}{x-1}+\frac{bx+c}{x^2-2x+2}=\frac{1}{(x-1)(x^2-2x+2)}$$

$$a(x^2-2x+2)+(bx+c)(x+1)=1$$

• Ops! I was tagging both variables with the same letter and trying to solve it :D Working on it right now! – Charlie Dec 10 '16 at 17:29