Is there any simpler way to find a remainder in multiple divisions? I got a question as follows.

$3x-5$ is the remainder when unknown $f(x)$ is divided by $x^2-x+1$
  that has relatively complicated  roots. Find the remainder when $f(x)$
  is divided by $(x^2-x+1)(x-1)$. Express your answer in terms of
  unknown $f(1)$, $x^2-x+1$, and $3x-5$.

Attempt
As the divisor is a cubic, then the remainder is at most a quadratic $a x^2 + bx +c$. Let $x_1$ and $x_2$ be the complicated root of $x^2-x+1$.
Now I have
\begin{align}
3x_1-5 &= a x_1^2 + bx_1 + c\\
3x_2-5 &= a x_2^2 + b x_2 +c \\
f(1) &= a + b +c
\end{align}
Finding $a$, $b$ and $c$ seems to be extremely tedious for me.
Question
Is there any simpler method to find the remainder in question?
Edit:
I have edited the quoted problem. The answer can be in terms of the divisor $x^2-x+1$ as well as the remainder $3x-5$. 

A possible way is as follows:


*

*$f(x) = (x^2-x+1)q(x) + 3x-5$

*$q(x) = (x-1)r(x) + c$
Hence,
$$f(x) = (x^2-x+1)((x-1)r(x) + c) + 3x -5$$ $$= (x^2-x+1)(x-1)r(x) + c(x^2-x+1) + 3x-5$$
Now you can find $c$:
$$f(1) =c -2 \Leftrightarrow c=2+f(1)$$
So, you get 
$$f(x) = (x^2-x+1)(x-1)r(x) + \color{blue}{(2+f(1))(x^2-x+1) + 3x-5}$$
I leave it up to you to collect like terms of the remainder as it suits you.
