We have the following recurrence:
$t_{n}=\begin{cases}n\\ 2t_{n-1}-2t_{n-3}+t_{n-4}\end{cases}$
if $n= 0$ or $1$ then $t_n = n$ else it's $2t_{n-1}-2t_{n-3}+t_{n-4}$.
The homogenous recurrence is $t_n - 2t_{n-1}+2t_{n-3}-t_{n-4} = 0$
The characteristic polynomial is $p(x) = x^4 -2x^3 + 2x - 1$
In the provided solutions for this exercise they verify if $p(1)$ and $p(-1)$ are $= 0$ before carrying with a long division of $p(x)$ with $(x^2 - 1)$. The reason for this is if $p(1)$ and $p(-1)$ are indeed roots (meaning causes $p(x)$ to equal $0$) then we have $p(x) = (x - 1)(x + 1)q(x) = (x^2 - 1)q(x)$.
Therefore we use long division to find $q(x)$ so we can find the remainding roots.
What is not clear to me is how did they known that 1 and -1 could potentially be roots before verifing it? Is it a common pattern for solving questions like these. Should I conclude that we always should check for 1 and -1 as roots. What if they're not roots, what do we do then?