Let ${\bf A}(x)$ be an $(n-1) \times n$ polynomial matrix and ${\bf b}(x)$ be a $1 \times n$ polynomial vector. Suppose that
$$ \det\begin{bmatrix} {\bf A}(x) \\ {\bf b}(x) \end{bmatrix}=0 $$
does not have a root inside the unit circle.
What condition must a $1 \times n$ constant vector ${\bf c}$ satisfy in order for
$$ \det\begin{bmatrix} {\bf A}(x) \\ {\bf b}(x) - {\bf c} \end{bmatrix}=0 $$ to have a root inside the unit circle?
I guess it is useful to use the fact that
$$ \det\begin{bmatrix} {\bf A}(x) \\ {\bf b}(x) - {\bf c} \end{bmatrix} = \det\begin{bmatrix} {\bf A}(x) \\ {\bf b}(x) \end{bmatrix}-\det\begin{bmatrix} {\bf A}(x) \\ {\bf c} \end{bmatrix},$$
which means, in particular, that ${\bf c}={\bf b}(x_0)$ for some $x_0$ inside the unit circle is a sufficient condition.
A necessary and sufficient condition is that there exists a vector $\bf v$ and some $x_0$ inside the unit circle such that ${\bf c} = {\bf b}(x_0)+{\bf vA}(x_0)$.
If this is the case then
$$ \det\begin{bmatrix} {\bf A}(x_0) \\ {\bf b}(x_0) - {\bf c} \end{bmatrix} = \det\begin{bmatrix} {\bf A}(x_0) \\ {\bf vA}(x_0) \end{bmatrix}=0.$$
Since
$$ \det\begin{bmatrix} {\bf A}(x) \\ {\bf b}(x) \end{bmatrix}=0 $$
does not have a root inside the unit circle, it follows that ${\bf A}(x)$ has full row rank for all $x$ inside the unit circle. Therefore, if the condition is not satisfied ${\bf b}(x) - {\bf c}$ will not be linearly dependent with the rows of ${\bf A}(x)$ for all $x$ in the unit circle, so that
$$ \det\begin{bmatrix} {\bf A}(x) \\ {\bf b}(x)-{\bf c} \end{bmatrix}= 0 $$
does not have an inside root.