The determinant left| begin{array}{ccc} 4 + x | vedclass

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(B) Let $\Delta = \left| \begin{array}{ccc} 4 + x^2 & -6 & -2 \ -6 & 9 + x^2 & 3 \ -2 & 3 & 1 + x^2 \end{array} ight|$.Applying the column operati

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$x^3$

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(B) Let $\Delta = \left| \begin{array}{ccc} 4 + x^2 & -6 & -2 \ -6 & 9 + x^2 & 3 \ -2 & 3 & 1 + x^2 \end{array} ight|$.Applying the column operations $C_1 	o C_1 + x C_3$ and $C_2 	o C_2 - 3 C_3$:$\Delta = \left| \begin{array}{ccc} 4 + x^2 - 2x & -6 & -2 \ -6 + 3x & 9 + x^2 & 3 \ -2 + x + x^3 & 3 - 3 - 3x^2 & 1 + x^2 \end{array} ight|$ is not the most efficient path.Instead,expand the determinant directly or use row/column properties.After simplification,the determinant evaluates to $\Delta = x^2(x^2 + 14)$.Since $\Delta = x^2(x^2 + 14)$,it is divisible by $x$,$x^2$,and $(14 + x^2)$.It is not divisible by $x^3$ or $x^5$.

The determinant $\left| \begin{array}{ccc} 4 + x^2 & -6 & -2 \\ -6 & 9 + x^2 & 3 \\ -2 & 3 & 1 + x^2 \end{array} \right|$ for $x \neq 0$ is not divisible by:

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