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(D) Let $\Delta = \left| {\begin{array}{ccc} 4 + {x^2} & -6 & -2 \ -6 & 9 + {x^2} & 3 \ -2 & 3 & 1 + {x^2} \end{array}} ight|$.Performing row oper

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${x^5}$

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(D) Let $\Delta = \left| {\begin{array}{ccc} 4 + {x^2} & -6 & -2 \ -6 & 9 + {x^2} & 3 \ -2 & 3 & 1 + {x^2} \end{array}} ight|$.Performing row operations: $R_1 	o R_1 + 2R_3$ and $R_2 	o R_2 - 3R_3$:$\Delta = \left| {\begin{array}{ccc} 4 + {x^2} - 4 & -6 + 6 & -2 + 2 + 2{x^2} \ -6 + 6 & 9 + {x^2} - 9 & 3 - 3 - 3{x^2} \ -2 & 3 & 1 + {x^2} \end{array}} ight| = \left| {\begin{array}{ccc} {x^2} & 0 & 2{x^2} \ 0 & {x^2} & -3{x^2} \ -2 & 3 & 1 + {x^2} \end{array}} ight|$.Taking ${x^2}$ common from $R_1$ and $R_2$:$\Delta = {x^4} \left| {\begin{array}{ccc} 1 & 0 & 2 \ 0 & 1 & -3 \ -2 & 3 & 1 + {x^2} \end{array}} ight|$.Expanding along $R_1$:$\Delta = {x^4} [1(1 + {x^2} + 9) - 0 + 2(0 + 2)] = {x^4}(10 + {x^2} + 4) = {x^4}(14 + {x^2})$.Since $\Delta = {x^4}(14 + {x^2}) = x \cdot {x^3} \cdot (14 + {x^2})$,it is divisible by $x$,${x^3}$,and $(14 + {x^2})$,but not by ${x^5}$.

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

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