1

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 $3x-5$.

Display Name
  • 2,715

1 Answers1

1

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.

trancelocation
  • 32,243