<s>[INST] Find roots of $f(x)=\left(x^2-1\right)^2\left(x^2-2x+1\right)$ I need to find the roots for the following function:
$$f(x)=\left(x^2-1\right)^2\left(x^2-2x+1\right)$$
I guess I already found one by doing this:
$$ f(x)=(x^2-1)^2(x^2-2x+1)\\
         (x^2-1)^2 = 0 | +1\\
          x^2 = 1      | √ \\
          x = 1 \\
          P₀₁(1|0)$$
But I could need some help, to find the second zero digit.
 [/INST] A good way to find zeroes of a function is to see if you can write it as a multiplication of pieces, $f(x) = (x-\alpha)\cdot (x-\beta)\cdot(x-\gamma)\dots\cdot(x-\omega)$  If you can, then $\alpha,\beta,\gamma\cdots$ are all roots of the equation.
For your function:
$f(x)=(x^2-1)^2\cdot(x^2-2x+1)$
$f(x) = ((x+1)(x-1))^2\cdot (x-1)\cdot(x-1)$
$f(x) = (x+1)(x+1)(x-1)(x-1)(x-1)(x-1)$
Going from the first line to the second, note that $x^2 - 1$ is the difference of squares and for $x^2-2x+1$ it is the square of $x-1$
Thus the two roots are $+1$ with multiplicity 4 and $-1$ with multiplicity 2
</s>