Application of Factor Theorem to Determinants

Theorem 7.3 (Factor Theorem for determinants). If every entry of a square matrix $A$ is a polynomial in $x$, and $|A|$ vanishes (equals $0$) at $x=a$, then $(x-a)$ is a factor of $|A|$.

This is exactly the ordinary polynomial Factor Theorem, applied to the determinant viewed as a polynomial in $x$. It is especially powerful when $|A|$ is in cyclic symmetric form — each row of $A$ obtained from the row above it by cycling the variables $a\to b\to c\to a$ — because then finding one factor by substitution automatically hands you the other two by symmetry.

Working rules:

1. If setting $x=b$ in place of $a$ makes two rows (or columns) of $|A|$ identical, then $|A|=0$ there (Property 4), so by the Factor Theorem $(a-b)$ is a factor.
2. If $r$ rows (or columns) become identical when $x=a$ is substituted (order $n\ge r$), then $(x-a)^{r-1}$ is a factor.
3. In a cyclic symmetric determinant, once $(a-b)$ is found to be a factor, $(b-c)$ and $(c-a)$ follow by the cyclic symmetry — no extra work needed.
4. Let $m$ = (degree of the product of the factors found so far) $-$ (degree of the product of the leading-diagonal entries of $|A|$). The remaining factor is then: a constant $k$ if $m=0$; $k(a+b+c)$ if $m=1$; $k(a^2+b^2+c^2)+l(ab+bc+ca)$ if $m=2$.
5. The constant(s) $k$ (and $l$) are found by comparing the coefficient of the highest power on both sides, or by substituting a couple of convenient numeric triples and solving the resulting equations. …