Let $K$ be a subfield of $R$ (e.g., $K=\mathbb Q$) and $f\in K[x]$ be a monic polynomial of degree $n$ all of whose roots are real. We want to compute a symmetric matrix $A\in K^{n\times n}$ such that $f=\det(xI_n-A)$. By squarefree decomposition in $K[x]$ we can assume that $f$ has only simple roots (otherwise perform euclidean divisions to compute the squarefree decomposition, find a determinantal representation for each factor in this decomposition and use diagonal blocks).
Of course, the idea is to start off with the companion matrix $C\in K^{n\times n}$ of $f$ which represents the vector space endomorphism $$\varphi\colon K[x]/(f)\to K[x]/(f),\ \overline p\mapsto\overline{xp}$$ with respect to the canonical monomial basis. Now $C$ itself is almost never symmetric but $\varphi$ is self-adjoint with respect to the $L_2$ scalar product given by a measure whose mass is uniformly distributed on the roots of $f$. Denoting the usual scalar product on $\mathbb R^n$ by $\langle.,.\rangle$ and by $V\in \mathbb R^{n\times n}$ the Vandermonde matrix corresponding to the roots of $f$, this means that $\langle VCx,Vy\rangle=\langle Vx,VCy\rangle$ for all $x,y\in\mathbb R^n$. In other words, we have $C^TV^TV=V^TVC$.
Now the Hermite matrix $H:=V^TV$ of $f$ comes naturally into play. It is a Hankel matrix whose entries are power sums of roots of $f$. But these power sums are traces of powers of $C$ and therefore lie in $K$ and can easily be computed. The idea is to replace $V$ by another matrix $W$ with $H=W^TW$. In contrast to $V$, the matrix $W$ should have entries in $K$ and should be easily computed. But this is possible: Just take any $W\in K^{n\times n}$ with $H=W^TW$. By Cholesky decomposition, you get such a triangular $W$.
Now $C^TW^TW=C^TH=C^TV^TV=V^TVC=HC=W^TWC$ and since $W$ is invertible (note that $V$ and $H$ are invertible since all roots of $f$ are simple) this shows that $WCW^{-1}$ is symmetric. So you can set $A:=WCW^{-1}$.
All this is folklore.
If you want a tridiagonal $W$ then you would have to perform Sturm's algorithm as indicated above by Denis. In fact, finding a tridiagonal $W$ is essentially equivalent to Sturm's algorithm, see the recent interesting work of Ronan Quarez: