Here is a simple argument showing that you can get any polynomial up to a constant factor. As Rob Israel and my comments above show, you might not be able to get rid of that constant.

If $\det(A_1 t + B_1) = f_1(t)$ and $\det(A_2 t + B_2) = f_2(t)$, then $\det \left( \begin{smallmatrix} A_1 t+B_1 & 0 \\ 0 & A_2t + B_2 \end{smallmatrix} \right) = f_1(t) f_2(t)$. Thus, we are immediately reduced to the case that $f$ is irreducible.

We also may ignore the case that $f(t) = t$, as that is easy to achieve.

Let $k$ be our ground field, and let $L = k(t)/f(t)$. Since $f$ is irreducible, this is a field. Choose a nonzero $k$-linear map $\tau: L \to k$. (If our extension is separable, it would be elegant to choose the trace, but this choice doesn't actually matter.) Let $u \in L$ be such that $\tau(u) \neq 0$.

Define two $k$-linear symmetric bilinear forms on $L$: $\langle x,y \rangle_1 = \tau(xy)$ and $\langle x,y \rangle_2 = \tau(xyt)$. These are symmetric because multiplication is commutative. The first is non-degenerate because, for any non-zero $x \in L$, we have $\langle x, u x^{-1} \rangle_1 \neq 0$. A similar argument applies to $\langle \ , \ \rangle_2$, using that $t \neq 0$. (This is why we had to exclude $f(t)=t$.)

Observe that $$\langle x,y \rangle_2 = \langle x, T y \rangle_1$$ where $T$ is the linear map $L \to L$ given by multiplication by $t$. Choose a $k$-basis for $L$. Let $A$ and $B$ be the matrices of $\langle \ , \ \rangle_1$ and $\langle \ , \ \rangle_2$ in this basis. Then the above equation shows that $A = BT$.

Since our bilinear forms are symmetric, $A$ and $B$ are symmetric matrices. Since our bilinear forms are nondegenerate, they are invertible. Since $A = BT$, we have $\det(A \lambda+B) = \det(B) \det(\lambda T + \mathrm{Id})$. The characteristic polynomial of $T$ is $f(t)$, so we are done.

Here is a simple argument showing that you can get any polynomial up to a constant factor. As Rob Israel and my comments above show, you might not be able to get rid of that constant.

If $\det(A_1 t + B_1) = f_1(t)$ and $\det(A_2 t + B_2) = f_2(t)$, then $\det \left( \begin{smallmatrix} A_1 t+B_1 & 0 \\ 0 & A_2t + B_2 \end{smallmatrix} \right) = f_1(t) f_2(t)$. Thus, we are immediately reduced to the case that $f$ is irreducible.

We also may ignore the case that $f(t) = t$, as that is easy to achieve.

Let $k$ be our ground field, and let $L = k(t)/f(t)$. Since $f$ is irreducible, this is a field. Choose a nonzero $k$-linear map $\tau: L \to k$. (If our extension is separable, it would be elegant to choose the trace, but this choice doesn't actually matter.) Let $u \in L$ be such that $\tau(u) \neq 0$.

Define two $k$-linear symmetric bilinear forms on $L$: $\langle x,y \rangle_1 = \tau(xy)$ and $\langle x,y \rangle_2 = \tau(xyt)$. These are symmetric because multiplication is commutative. The first is non-degenerate because, for any non-zero $x \in L$, we have $\langle x, u x^{-1} \rangle_1 \neq 0$. A similar argument applies to $\langle \ , \ \rangle_2$, using that $t \neq 0$. (This is why we had to exclude $f(t)=t$.)

Observe that $$\langle x,y \rangle_2 = \langle x, T y \rangle_1$$ where $T$ is the linear map $L \to L$ given by multiplication by $t$. Choose a $k$-basis for $L$. Let $A$ and $B$ be the matrices of $\langle \ , \ \rangle_1$ and $\langle \ , \ \rangle_2$ in this basis. Then the above equation shows that $A = BT$.

Since our bilinear forms are symmetric, $A$ and $B$ are symmetric matrices. Since our bilinear forms are nondegenerate, they are invertible. Since $A = BT$, we have $\det(A \lambda+B) = \det(B) \det(\lambda T + \mathrm{Id})$. The characteristic polynomial of $T$ is $f(t)$, so we are done.

ADDED: It looks like Allison Miller has just about the same argument, and also has some references to where this idea can be found in the literature.