Zeros are a good thing, as they mean there is no contribution from the cofactor there.
$$
\det A =
1 \cdot (-1)^{1 + 1} \det S_{11} + 2 \cdot (-1)^{1+2} \det S_{12} + 0 \cdot \dotsb + 0 \cdot \dotsb
$$
with
$$
S_{11} =
\begin{pmatrix}
\times & \times & \times & \times \\
\times & 4 & 0 & 0 \\
\times & 0 & 5 & 6 \\
\times & 0 & 7 & 8
\end{pmatrix}
=
\begin{pmatrix}
4 & 0 & 0 \\
0 & 5 & 6 \\
0 & 7 & 8
\end{pmatrix}
\\
S_{12} =
\begin{pmatrix}
\times & \times & \times & \times \\
3 & \times & 0 & 0 \\
0 & \times & 5 & 6 \\
0 & \times & 7 & 8
\end{pmatrix}
=
\begin{pmatrix}
3 & 0 & 0 \\
0 & 5 & 6 \\
0 & 7 & 8
\end{pmatrix}
$$
where $S_{ij}$ is the matrix $A$ with row $i$ and column $j$ removed.
The determinants of $S_{11}$ and $S_{12}$ are then calculated again by expansion along the first row, e.g.
$$
\det S_{11} =
4 \cdot (-1)^{1+1} \det
\begin{pmatrix}
\times & \times & \times \\
\times & 5 & 6 \\
\times & 7 & 8
\end{pmatrix}
=
4 \det
\begin{pmatrix}
5 & 6 \\
7 & 8
\end{pmatrix}
\\
\det S_{12} =
3 \cdot (-1)^{1+1} \det
\begin{pmatrix}
\times & \times & \times \\
\times & 5 & 6 \\
\times & 7 & 8
\end{pmatrix}
=
3 \det
\begin{pmatrix}
5 & 6 \\
7 & 8
\end{pmatrix}
$$
until one hits a $2\times2$ matrix where one knows the direct formula.