I do not pretend that the solution I propose is simpler, but its big advantage is that it is "computer oriented", therefore able to manage any polynomial degree.
Let us first give a name to the initial polynomial :
$$P(x)=x^4+12x-5$$
We are going to use the resultant of two monic polynomials $P$ and $Q$, which is defined as the product of all the differences between their roots.
$$
\operatorname{Res}(P,Q)=\prod(\alpha_i-\beta_j),
$$
($\alpha_i :$ roots of $P$, $\beta_j :$ roots of $Q$).
$\operatorname{Res}(P,Q)$ is zero if and only if $P$ and $Q$ have a common root.
The interest of resultants is mainly in issues like this one with presence of parameters. Here, we are going to introduce a parameter $s$ by taking the resultant of the initial polynomial $P$ and the new polynomial
$$Q_s(x):=P(s-x)$$
$\operatorname{Res}(P,Q_s)$ will be a polynomial in the variable $s$ which will be zero if and only if there is a value of $s$ such that
$$\alpha_i=s-\beta_j \ \ \ \iff \ \ \ s=\alpha_i+\beta_j$$
for some $i,j$, which is what we desire.
An explicit form of $Q_s$ is :
$$Q_s(x)=x^4 + \underbrace{(-4s)}_{A}x^3 + \underbrace{(6s^2)}_{B}x^2 + \underbrace{(- 4s^3 - 12)}_{C}x + \underbrace{(s^4 + 12s - 5)}_{D}\tag{1}$$
Let us now form the resultant matrix of $P$ and $Q_s$ (obtained by repeating 4 times the coefficients of the first, then the second polynomial, with a shift at each new row as indicated in the reference given upwards) :
$$R=\left(\begin{array}{cccccccc} 1& 0& 0& 12& -5& 0& 0& 0\\
0& 1& 0& 0& 12& -5& 0& 0\\
0& 0& 1& 0& 0& 12& -5& 0\\
0& 0& 0& 1& 0& 0& 12& -5\\
1& A& B& C& D& 0& 0& 0\\
0& 1& A& B& C& D& 0& 0\\
0& 0& 1& A& B& C& D& 0\\
0& 0& 0& 1& A& B& C& D
\end{array}\right)
$$
Let us expand and factorize $\det(R)$ (all operations done with a Computer Algebra System) :
$$\det(R)=(s^2 + 4s - 4)(s^2 - 4s + 20)(\underbrace{(s - 2)(s + 2)(s^4 + 4s^2 + 36)}_{\color{red}{s^6+20s^2-144}})^2$$
The first two factors have to be discarded because they correspond to spurious roots $x_k+x_k$.
It remains the content of the square factor which is the looked for polynomial...
Here is the corresponding (Matlab) program :
function main;
syms s x; % symbolic letters
P=[1,0,0,12,-5]; % it's all we have to give ; the rest is computed...
lp=length(P);pol=0;
for k=1:lp;
pol=pol+P(k)*x^(lp-k);
end;
Qs=coeffs(collect(expand(subs(pol,x,s-x)),x),x);
Qs=fliplr(Qs); % list reversal ("flip left right')
R=Resu(P,Qs)'
factor(det(R))
%
function R=Resu(P,Q) ; % Resultant matrix
p=length(P)-1;q=length(Q)-1; % degrees of P,Q
R=sym(zeros(p+q));
for k=1:q
R(k,k:k+p)=P; % progressive shifting
end
for k=1:p
R(k+q,k:k+q)=Q;
end
R=R'
Remarks :
1) The presence of a square around the solution isn't in fact surprizing : we have the same phenomenon with the discriminant of a polynomial :
$$
\operatorname{Disc}(P)=\operatorname{Res}(P,P')=\prod_{i \neq j}(\alpha_i-\alpha_j)^2,
$$
2) A similar issue can be found here.
3) In (1), if necessary, these coefficients $A,B,C$ and $D$ can be considered as issued from a Taylor expansion: $D=P(s), C=-P'(s), B=\tfrac12 P''(s), A=-\tfrac16 P'''(s)$.
4) Another category of problems, polynomial transformations, for example finding the polynomial whose roots are the $\alpha_k+1/\alpha_k$ where the $\alpha_k$s are the roots of a given polynomial $P$ can as well be solved using resultants see here.