This question already has an answer here:
I assume $a\neq 0$, for otherwise the question is trivial.
Note this is the characteristic polynomial
$$
p_A(x)=\det(xI_4-A)
$$
of the real symmetric matrix
$$A=\pmatrix{0&-a&-a&-a\\ -a&0&-a&-a\\-a&-a&0&-a\\-a&-a&-a&0}.$$
We know such a matrix is diagonalizable via an orthogonal matrix. In particular, its spectrum is all real and its characteristic polynomial splits over $\mathbb{R}$.
For $x=a$, we get a rank one matrix $aI_4-A$. This means that $a$ is an eigenvalue and that the corresponding eigenspace has dimension $3$, which is also the multiplicity of $a$ in the characteristic polynomial.
There only remains one eigenvalue $\lambda$ of multiplicity $1$ to find. It suffices to take the trace:
$$\mbox{Tr} A=0=3a+\lambda \quad\Rightarrow\quad\lambda=-3a.$$
So
$$
p_A(x)=(x-a)^3(x+3a).
$$