= Degree drop in a specialized resultant
The actual-degree condition in the <resultant> theorem matters when <coefficients> depend on a parameter. Take
$$
f_t(x)=tx+1,\qquad g_t(x)=tx+2.
$$
The nominal degree-one <Sylvester matrix> is $\begin{pmatrix}t&1\\t&2\end{pmatrix}$, whose <determinant> is $t$. Its specialization at $t=0$ therefore vanishes. But the specialized <polynomials> are the nonzero constants $1$ and $2$, which have no <common root> at all.
For $t\ne0$, the <roots of a polynomial> are $-1/t$ and $-2/t$; neither has a finite <limit> as $t\to0$. Both <leading coefficients> vanish at the exceptional parameter. The fixed-size <determinant> has retained the original degree format even though the specialized <polynomial degrees> have dropped. Recompute using the actual specialized <polynomials> and check the original equations before accepting a zero of a parameter-dependent <resultant> as a solution.
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