I know a little bit about the resultant of two polynomials but I couldn't find any example whatsoever. So I was wondering If you could illustrate the process of finding the resultant of two polynomials for $f = x^3-2x+4$ and $g = x^5+x^3-2x+4$. Also If you could give me another examples that you might think it could help me understant this concept would be great.
What I've tried so far was to see if f and g have common roots in which case the resultant would be 0, but that didn't work. After that I have tried to calculate the determinant of the Sylvester matrix but I'm not sure how to calculate that determinant:
$$\begin{pmatrix}
1 & 0 & 1 & -1 & 0 & 4 &0 &0 \\
0 & 1 & 0 & 1 & -1 & 0 & 4 & 0\\
0 & 0 & 1 & 0 & 1 & -1 & 0 & 4\\
1 & 0 & -2& 4& 0& 0 & 0 & 0\\
0 & 1 & 0& -2 & 4& 0& 0 &0 \\
0 & 0 & 1 & 0 & -2 & 4 & 0 &0 \\
0& 0 & 0 & 1 & 0 & -2 & 4 & 0\\
0 & 0 & 0 & 0 & 1 & 0 & -2 & 4
\end{pmatrix}$$
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related: https://math.stackexchange.com/q/1048049/173147 – glS Apr 27 '21 at 15:30
2 Answers
Your matrix would be correct if $g$ were $x^5+x^3-x^2+4$. But since $g$ is $x^5+x^3-2x+4$ instead, it should be $$\begin{pmatrix} 1 & 0 & 1 & 0 & -2 & 4 &0 &0 \\ 0 & 1 & 0 & 1 & 0 & -2 & 4 & 0\\ 0 & 0 & 1 & 0 & 1 & 0 & -2 & 4\\ 1 & 0 & -2& 4& 0& 0 & 0 & 0\\ 0 & 1 & 0& -2 & 4& 0& 0 &0 \\ 0 & 0 & 1 & 0 & -2 & 4 & 0 &0 \\ 0& 0 & 0 & 1 & 0 & -2 & 4 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & -2 & 4 \end{pmatrix}$$ instead.
You can compute the determinant as you would compute the determinant of any other matrix. There seems to be a faster algorithm in §2.3.3 of M. Pohst and H. Zassenhaus, Algorithmic algebraic number theory, CUP 1993. This may and may not be the same algorithm that was discussed in How does one compute a resultant by using Euclidean algorithm? (I am not an expert in this subject).
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What this Sylvester's determinant means is a necessary and sufficient condition for both polynomials $f(x)=x^3-2x+4$ and $g(x)=x^5 +x^3-2x+4$ have common roots which is not the case because the roots of $f$ are $-2$ and $1\pm i$ and none of these three roots is a root of $g$.
So we can assure whitout calculation that the value of this determinant is distinct of zero. But this determinant of Sylvester does not give a method to find the resultant.
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