A way to split this determinant as a product of two? The question asks if the det of the following 3*3  matrix is divisible by $$ x^4,x^3,x^2,x $$
$$
\begin{bmatrix}
a^2+x^2 & ab & ac \\
ab & b^2+x^2 & bc \\
ac & bc & c^2+x^2 \\
\end{bmatrix}
$$
I just shovelled through using the standard method and got $$ x^4(b^2+c^2+a^2+x^2) $$ So the answer is "Yes" but is there a simpler way like splitting the det as a product of two dets? If not that is there anything else?

The matrix can be written as
$$A = x^2I_3 + \begin{bmatrix}a\\b\\c\end{bmatrix} \begin{bmatrix}a & b & c\end{bmatrix}$$
Now by Sylvester determinant theorem, we have
\begin{align}
\det(A) & = x^6 \det\left(I_3 + \dfrac1{x^2}\begin{bmatrix}a\\b\\c\end{bmatrix} \begin{bmatrix}a & b & c\end{bmatrix}\right) = x^6 \left(1+\dfrac1{x^2}\begin{bmatrix}a & b & c\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}\right)\\
& = x^4(x^2+a^2+b^2+c^2)
\end{align}
