I recently started learning the applications of the resultant and I found the following statement:
Let $f(x),g(x) \in \mathbb{K}[x]$ (where $\mathbb{K}$ is some field) and $deg(f) \geq deg(g)$ prove that $Res(f+gh,g) = Res(f,g)$ where $h(x) \in \mathbb{K}[x]$ with $deg(h)\leq deg(f)-deg(g)$
($Res$ here is the resultant - that we compute as the determinant of the Sylvester matrix, also note that $\mathbb{K}$ isn't necessarily algebraically closed).
any help would be appreciated, thanks in advance
The key here is to remember that the determinant is invariant under adding multiples of one row to another row. Consider the following example - let $f=x^3+2x^2+3x+4$, $g=x+1$, and $h=x^2+3$. Then the Sylvester matrix asociated to $Res(f,g)$ is $$\begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix} $$
while the Sylvester matrix associated to $Res(f+gh,g)$ is $$\begin{pmatrix} 2 & 3 & 6 & 7 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix} $$ which can be obtained from the first matrix by adding the 2nd row to the 1st and three times the 4th row to the first. (Adding one times the second row corresponds to the coefficient of $x^2$ being $1$, adding zero times the third row to the first row corresponds to the coefficient of $x$ being zero, and three times the fourth row corresponds to the constant term being $3$.)
The general case is completely similar, except you will need to do this for every line which has the coefficients of $f$ on it. It is immediately clear that you can do this based on the construction of the Sylvester matrix, and by the invariance of the determinant under adding multiples of one row to another, this will always give you the same answer.