Determine the multiplicity of a root in determinant Consider the determinant
$$\det(A(x,z))=\begin{vmatrix}1-x&1&1&1\\1&1+x&1&1\\1&1&1-z&1\\1&1&1&1+z\end{vmatrix}.$$
It is easy to see that if we write the determinant as a bivariate polynomial $p(x,z)$, then $x=0$ and $z=0$ are its roots, i.e. $p(0,z)=p(x,0)=0$. But how can we show that these roots have multiplicity 2?
Specifically, I'm interested if it is possible to determine the multiplicity of a root in a determinant without computing the determinant explicitely. One possibility would be to use the Jacobi formula to compute, e.g., $\frac{\partial}{\partial x}\det(A(x,z))$, but this task seems to be even more complicated. Probably there are some tricks that allow us to make a conclusion about the multiplicity of a root with less effort?
EDIT: The solution is supposed to follow this scheme:

*

*Show that $x=0$ ($z=0$) is a root of the polynomial.

*Show that $x=0$ ($z=0$) is a double root of the polynomial.

*Deduce that $p(x,z)=x^2z^2$.

It is the second item, where I have a problem.
 A: Adding or subtracting one row to another does not change the value of the determinant. So $$\det\begin{bmatrix}1-x&1&1&1\\1&1+x&1&1\\1&1&1-z&1\\1&1&1&1+z\end{bmatrix}=\det\begin{bmatrix}1&1&1&1+z\\0&x&0&-z\\0&0&z&z\\0&0&x&x+xz\end{bmatrix}=x^2z^2$$
NB It is $x=0$ and $z=0$ that are the roots of the polynomial.

EDIT: Let's expand $\det(A(x))$ as a series in $x$:
$\det(A(0))=0$ since it contains a repeated column of $1$s. This shows that $x=0$ is a root of $p(x,z)$.
To get the first order terms in $x$, recall that determinants are linear in each column separately, hence expanding the first and second columns gives \begin{align}\det(A(x))&=\det\begin{bmatrix}1&1&1&1\\1&1+x&1&1\\1&1&1-z&1\\1&1&1&1+z\end{bmatrix}-\det\begin{bmatrix}x&1&1&1\\0&1+x&1&1\\0&1&1-z&1\\0&1&1&1+z\end{bmatrix}\\
&=\det\begin{bmatrix}1&0&1&1\\1&x&1&1\\1&0&1-z&1\\1&0&1&1+z\end{bmatrix}-\det\begin{bmatrix}x&1&1&1\\0&1&1&1\\0&1&1-z&1\\0&1&1&1+z\end{bmatrix}-\det\begin{bmatrix}x&0&1&1\\0&x&1&1\\0&0&1-z&1\\0&0&1&1+z\end{bmatrix}\\
&=-x^2\det\begin{bmatrix}1-z&1\\1&1+z\end{bmatrix}\end{align} In the second equation, the first two determinants cancel out by noticing that the second can be obtained from the first by swapping the first two columns and rows. It is this cancellation that annihilates the $x$ term and produces the double root $x^2$.
