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.

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