Solve by factoring, then perform synthetic division. $${x^4 - 3x^3 + 3x^2 - x = 0}$$
$${x(x^3 - 3x^2 + 3x - 1)}$$
Now, we have to perform synthetic division, if I'm correct we get the 1 to divide by from the X outside of the parentheses from the second equation, which is above this paragraph, and is also the factored form of the first equation.
$$\frac{x - 3 + 3 - 1}{1}$$
The answer is ${x = 0, 1}$. Does the 0 come from the remainder, and do the 1 come from the divisor?
 A: Synthetic division is not the greatest tool, in my opinion, because it makes the division process very mysterious. It's a shorthand for long division, and I would recommend writing out the steps of long division instead.
Anyway, to use synthetic division here, you can notice by plugging in a few values of $x$ that $x=1$ makes the cubic factor disappear: $1^3 - 3(1)^2 + 3(1) - 1 = 0$. On to synthetic division, with $1$ as the magic number:
$$\begin{array}{c|cccc}
  & 1 & -3 & 3 & -1\\
1 &   & 1 & -2 & 1\\
\hline
  & 1 & -2 & 1 & 0
\end{array}$$
This means that $x^3 - 3x^2 + 3x - 1 = (x-1)(x^2 - 2x +1)$. (There is no remainder term since the last number in the bottom right is $0$. For that matter, we knew in advance that there would be no remainder since $x=1$ is a zero.)
Now you could repeat the process with $x^2-2x+1$, but hopefully you recognize that $x^2-2x+1 = (x-1)^2$. Therefore, we have $x^3 - 3x^2 + 3x - 1 = (x-1)^3$. Coming back to the original polynomial of degree $4$, we can see that:
$$x^4 - 3x^3 + 3x^2 - x = x(x^3 - 3x^2 + 3x - 1) = x(x-1)^3.$$
The solutions $x=0,x=1$ come from the factors in this last expression ($x$ and $x-1$, respectively).
A: $$x^3-3x^2+3x-1=x^3-x^2-2x^2+2x+x-1=x^2(x-1)-2x(x-1)+(x-1)=(x-1)(x^2-2x+1)=(x-1)^3$$
