linear algebra - Does the inverse of a polynomial matrix have polynomial growth?

Let $M : \mathbb{R}^n \to \mathbb{R}^{n \times n}$ be a matrix-valued function whose entries $m_{ij}(x_1, \dots, x_n)$ are all multivariate polynomials with real coefficients. Suppose that $M(\mathbf{x})$ is invertible for every $\mathbf{x} \in \mathbb{R}^n$. I would like to show that $M^{-1}$ has at most polynomial growth.

That is, let $\|\cdot\|$ be your favorite matrix norm on $\mathbb{R}^{n \times n}$. I want to show there are constants $C,N$ such that $\|M^{-1}(\mathbf{x})\| \le C (1 + |\mathbf{x}|)^N$.

One possible approach starts as follows. Let $p(\lambda, \mathbf{x}) = \det(M(\mathbf{x}) - I \lambda)$ be the characteristic polynomial of $M$. Then, if we consider $M$ as a matrix over the commutative ring $\mathbb{R}[\mathbf{x}] = \mathbb{R}[x_1, \dots, x_n]$, we can apply the Cayley-Hamilton theorem to conclude that $p(M(\mathbf{x}), \mathbf{x}) = 0$. Since $M(\mathbf{x})$ is invertible for every $\mathbf{x}$, we have that $p(0, \mathbf{x}) = \det(M(\mathbf{x}))$ is a polynomial with no zeros; let us suppose $p(0, \mathbf{x}) > 0$. Then by writing $p(\lambda, \mathbf{x}) = \lambda r(\lambda, \mathbf{x}) + p(0, \mathbf{x})$ for some other polynomial $r$, we get that $$M(\mathbf{x})^{-1} = -\frac{r(M(\mathbf{x}), \mathbf{x})}{p(0,\mathbf{x})}.$$ The numerator certainly has polynomial growth, so it would remain to estimate $1/p(0,\mathbf{x}) = 1/\det(M(\mathbf{x}))$. I am not sure of an easy way to do that; I considered applying Hilbert's 17th problem (as solved by Artin) to write $p(0,\mathbf{x})$ as a sum of squares of rational functions, but that seems much too complicated.

I feel like there must be something much simpler that I am missing.

Answer

David Speyer's answer here showed that the reciprocal of a polynomial has polynomial growth, thanks to Stengle's Positivstellensatz. So the approach described in the question works, and the answer to the question is positive.