If $x=-4$ is a root of $\begin{vmatrix}x&2&3\\1&x&1\\3&2&x\end{vmatrix}=0$, then the sum of the other 2 roots is

4

The correct answer is Option (1) → 4

Given determinant:

$\begin{vmatrix} x & 2 & 3 \\ 1 & x & 1 \\ 3 & 2 & x \end{vmatrix} = 0$

Use standard formula for 3×3 determinant:

$x(x \cdot x - 1 \cdot 2) - 2(1 \cdot x - 1 \cdot 3) + 3(1 \cdot 2 - x \cdot 3) = 0$

Simplify:

$x^3 - 2x - 2x + 6 + 6 - 9x = x^3 - 13x + 12 = 0$

Given root $x = -4$, factor it out:

Divide $x^3 - 13x + 12$ by $(x + 4)$ → quotient: $x^2 - 4x + 3$

Factor $x^2 - 4x + 3 = 0$ → $(x - 1)(x - 3) = 0$

The other roots are $x = 1$ and $x = 3$

Sum of other two roots: $1 + 3 = 4$