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(D) Given the determinant equation: $\left| \begin{array}{ccc} x & 3 & 7 \ 2 & x & -2 \ 7 & 8 & x \end{array} ight| = 0$. Expanding along the firs

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$2$ and $-7$

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(D) Given the determinant equation: $\left| \begin{array}{ccc} x & 3 & 7 \ 2 & x & -2 \ 7 & 8 & x \end{array} ight| = 0$. Expanding along the first row: $x(x^2 - (-16)) - 3(2x - (-14)) + 7(16 - 7x) = 0$ $x(x^2 + 16) - 3(2x + 14) + 7(16 - 7x) = 0$ $x^3 + 16x - 6x - 42 + 112 - 49x = 0$ $x^3 - 39x + 70 = 0$. Since $x = 5$ is a root,$(x - 5)$ must be a factor. Performing polynomial division or synthetic division on $x^3 - 39x + 70 = 0$: $x^3 - 5x^2 + 5x^2 - 25x - 14x + 70 = 0$ $x^2(x - 5) + 5x(x - 5) - 14(x - 5) = 0$ $(x - 5)(x^2 + 5x - 14) = 0$ $(x - 5)(x + 7)(x - 2) = 0$. Thus,the roots are $x = 5, 2, -7$. The other two roots are $2$ and $-7$.

If $5$ is one root of the equation $\left| \begin{array}{ccc} x & 3 & 7 \\ 2 & x & -2 \\ 7 & 8 & x \end{array} \right| = 0$,then the other two roots of the equation are:

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