<s>[INST] Quickest way to find characteristic polynomial from a given matrix Find the Rational form of 
$$
A=\begin{pmatrix} 1&2&0&4\\ \:\:\:4&1&2&0\\ \:\:\:0&4&1&2\\ \:\:\:2&0&4&1\end{pmatrix}
$$
I don't wanna the solution, instead I would like to know a quickest  way to calculate $\det (\lambda I-A)$. 
$$
\begin{vmatrix}
x-1 &-2  &0  &-4 \\ 
 -4&x-1  &-2  &0 \\ 
 0&-4  &x-1  & -2 \\ 
 -2&0  &-4  &x-1 
\end{vmatrix}
$$
here I could see the polynomial but the procedure is quite long.
https://www.symbolab.com/solver/matrix-eigenvectors-calculator/eigenvectors%20%5Cbegin%7Bpmatrix%7D1%262%260%264%5C%5C%20%20%20%20%204%261%262%260%5C%5C%20%20%20%20%200%264%261%262%5C%5C%20%20%20%20%202%260%264%261%5Cend%7Bpmatrix%7D
 [/INST] Add columns $\,2,3,4\,$ to the first one (this is a circulant matrix), and get:
$$\det(t I-A)=\begin{vmatrix}t-7&-2&0&-4\\
t-7&t-1&-2&0\\
t-7&-4&t-1&-2\\
t-7&0&-4&t-1\end{vmatrix}\stackrel{R_i-R_1}=\begin{vmatrix}t-7&-2&0&-4\\
0&t+1&-2&4\\
0&-2&t-1&2\\
0&2&-4&t+3\end{vmatrix}=$$$${}$$
$$=(t-7)\begin{vmatrix}
t+1&-2&4\\
-2&t-1&2\\
2&-4&t+3\end{vmatrix}=(t-7)\left[(t^2-1)(t+3)+24+8t+8-4t-12-8t+8\right]=$$
$$=(t-7)\left[(t^2-1)(t+3)-4t+28\right]=(t-7)\left[t^3+3t^2-5t+25\right]=$$
$$=(t-7)(t+5)(t^2-2t+5)=(t+t)(t+5)(t-(1-2i))(t-(1+2i))$$
</s>