<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] With respect to The Case of the Missing Root:
In general,  we have $$x^2 = a^2 \iff x= \pm\sqrt {a^2} \iff x = \pm a$$
In this case, then $$x^2= 1 \iff x = \pm \sqrt 1 \iff x=\pm 1$$
So yes, $x = 1$ is one solution to $x^2 = 1$, and the other is $x=-1$. It never hurts to go back to check out your solutions by "plugging them into" the original equation for verification.
Tip: One way to remember to include both positive and negative roots of a quadratic such as this is to recognize when you have a difference of squares:
$$\begin{align} 
x^2 = a^2 &\iff x^2 - a^2 = 0\tag{difference of squares}\\ \\ 
& \iff (x+a)(x-a) = 0 \\ \\ 
&\iff x=-a\;\;\text{or}\;x =a
\end{align}$$

Now, putting it all together, we have 
$$\begin{align} f(x)& =\left(x^2-1\right)^2\left(x^2-2x+1\right)\\\\
& = \Big((x+1)(x-1)\Big)^2\left(x-1\right)^2\\\\
&= (x+1)^2(x-1)^4\\\\
f(x) &= (x+1)^2(x-1)^4 = 0 \implies x= -1\, \text{ or }\, x = 1.
\end{align}$$
The multiplicity of each root can be found by looking at the exponents: Root $x=1$ has multiplicity four, and root $x = -1$ has multiplicity two.
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