https://math.stackexchange.com/questions/1090570/factorise-a-matrix-using-the-factor-theorem#comment6735204_1091452. Can anyone please answer this question? It's actually a comment on an answer.
What you did is correct. But there is an easier way. Remember that for polynomial $p(x)$, if $p(a)=0$ then $(x-a)$ is a factor of $p(x)$.
Denote the determinant by $\Delta$. It is obviously a polynomial in $x,\ y$ and $z$. Now, note that:
$x=0\implies \Delta = 0$, so $x$ is a factor of $\Delta...
What about this matrix. $$\begin{vmatrix}? 1&1&1\\ x^2&y^2&z^2\\ x^3&y^3&z^3\\ \end{vmatrix}$$ x=1 is a factor by your method, but not really. — hyphen33 mins ago
What you did is correct. But there is an easier way. Remember that for polynomial $p(x)$, if $p(a)=0$ then $(x-a)$ is a factor of $p(x)$.
Denote the determinant by $\Delta$. It is obviously a polynomial in $x,\ y$ and $z$. Now, note that:
$x=0\implies \Delta = 0$, so $x$ is a factor of $\Delta...
Diving multivariable polynomials is hard. The factor theorem in the form that is used still holds though. You can consider $K[X_1,...,X_n]$ as $K[X_1,...,X_{n-1}][X]$ and apply the usual factor theorem.
What you did is correct. But there is an easier way. Remember that for polynomial $p(x)$, if $p(a)=0$ then $(x-a)$ is a factor of $p(x)$.
Denote the determinant by $\Delta$. It is obviously a polynomial in $x,\ y$ and $z$. Now, note that:
$x=0\implies \Delta = 0$, so $x$ is a factor of $\Delta...
Linear & Abstract algebra
Linear & Abstract algebra