We assume $a^2+av+u=0$ for $u,v\in\bar\Q$ and $u^2+uq+p=0$, $v^2+sv+r=0$ for $p,q,r,s\in\Q$. Find some monic polynomial $P\in\Q[X]$ such that $P(a)=0$.
$E$ is the subspace generated by $1,a,u,au,v,av,uv,auv$; we assume this is a basis of the eight dimensional space $E$. Multiplication by $a$ maps these vectors onto the following vectors:
\begin{eqnarray*}
1&\to&a\\
a&\to&a^2=-av-u\\
u&\to&au\\
au&\to&a^2u=-auv-u^2=-auv+qu+p1\\
v&\to&av\\
av&\to&a^2v=-av^2-uv=sav+ra-uv\\
uv&\to&auv\\
auv&\to&a^2uv=-auv^2-u^2v=au(sv+r)+(qu+p)v=sauv+rau+quv+pv
\end{eqnarray*}
Thus the matrix of this map with respect to the basis
$$
1,a,u,au,v,av,uv,auv
$$
is the transpose of
$$
\left(\begin{array}{cccccccc}
0&1&0&0&0&0&0&0\\
0&0&-1&0&0&-1&0&0\\
0&0&0&1&0&0&0&0\\
p&0&q&0&0&0&0&-1\\
0&0&0&0&0&1&0&0\\
0&r&0&0&0&s&-1&0\\
0&0&0&0&0&0&0&1\\
0&0&0&r&p&0&q&s
\end{array}\right)
$$
$a$ is a root of the characteristic polynomial $P\in\Q[X]$ of this matrix.
A=matrix(SR,[[0,1,0,0,0,0,0,0],[0,0,-1,0,0,-1,0,0],[0,0,0,1,0,0,0,0],[p,0,q,0,0,0,0,-1],[0,0,0,0,0,1,0,0],[0,r,0,0,0,s,-1,0],[0,0,0,0,0,0,0,1],[0,0,0,r,p,0,q,s]])
A.characteristic_polynomial()
This will give you
\begin{eqnarray*}
P&=&X^8
- 2sX^7
+ (s^2 - 2q + 2r)X^6
+ (3qs - 2rs)X^5\\
&&+ (-qs^2 + q^2 - 2qr + r^2 + 2p)X^4
+ (-q^2s + qrs - 2ps)X^3\\
&&+ (q^2r + ps^2 - 2pq - 2pr)X^2
+ pqsX
+ p^2
\end{eqnarray*}
For $p,q,r,s\in\Z$ we have: $P\in\Z[X]$ and by Gauß' lemma $P$ is irreducible over $\Q[X]$ iff it's irreducible over $\Z[X]$. In order to prove irreducibility of $P$ you may utilize Eisenstein's criterion or irreducibility of $P$ over some $\Z_n$ for some. In particular for $p=2,q=0,r=3,s=0$, i.e. $u=\sqrt2$, $v=\sqrt3$ and $a^2+\sqrt3 u+\sqrt2=0$:
$$
P=X^8 + 6*X^6 + 13*X^4 - 12*X^2 + 4
$$
To check irreducibility:
R=PolynomialRing(ZZ,'X')
X=R.gen()
P=X^8 + 6*X^6 + 13*X^4 - 12*X^2 + 4
P.is_irreducible()
P.factor()
which shows that $P$ is irreducible.