If $x = 2$ is a root of $\det\left[\begin{smallmatrix}x&-6&-1\\2&-3x&x-3\\-3&2x&x+2\end{smallmatrix}\right]=0$, find other two roots

Question

If $x = 2$ is a root of equation $$ \begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3\\ -3 & 2x & x+2 \end{vmatrix} = 0 $$ Then find the other two roots.

I solved it and got a cubic equation, and then I divided it by $(x-2)$ to get the other two roots. But this is a long method to do.

Please help me with some shorter approach to this question.

Answer 1

$$ \begin{align*} \begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3\\ -3 & 2x & x+2 \end{vmatrix} & = \begin{vmatrix} x-2 & 3x-6 & 2-x \\ 2 & -3x & x-3\\ -3 & 2x & x+2 \end{vmatrix} \\ & = (x-2) \begin{vmatrix} 1 & 3 & -1 \\ 2 & -3x & x-3\\ -3 & 2x & x+2 \end{vmatrix} \\ & = (x-2) \begin{vmatrix} 1 & 0 & 0 \\ 2 & -3x-6 & x-1\\ -3 & 2x+9 & x-1 \end{vmatrix} \\ & = (x-2)(x-1) \begin{vmatrix} 1 & 0 & 0 \\ 2 & -3x-6 & 1\\ -3 & 2x+9 & 1 \end{vmatrix} \\ & = (x-2)(x-1) \begin{vmatrix} 1 & 0 & 0 \\ 5 & -5x-15 & 0\\ -3 & 2x+9 & 1 \end{vmatrix} \\ & = (x-2)(x-1)(-5x-15) \\ & = -5(x-2)(x-1)(x+3) \\ \end{align*} $$

Therefore, the other roots are $1$ and $-3$.

Answer 2

$\begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3\\ -3 & 2x & x+2 \end{vmatrix} = 0$

$\begin{vmatrix} x-2 & 3x-6 & 2-x\\ 2 & -3x & x-3\\ -3 & 2x & x+2 \end{vmatrix} = 0$ $R_1 \rightarrow R_1-R_2$

$(x-2)\begin{vmatrix} 1 & 3 & -1\\ 2 & -3x & x-3\\ -3 & 2x & x+2 \end{vmatrix} = 0$ $\begin{vmatrix} 1 & 3 & -1\\ 0 & -3x-6 & x-1\\ 0 & 2x+9 & x-1 \end{vmatrix} = 0$ $R_2 \rightarrow R_2-2R_1$, $R_3 \rightarrow R_3+3R_1$

Now open the determinant using $C_1$ clearly,one factor is (x-1). You get -3x-6= 2x+9, x=-3.