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 
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.
 A: $$
\begin{align*}
\begin{vmatrix}
x & -6 & -1 \\
2 & -3x & x-3\\
-3 & 2x & x+2 
\end{vmatrix} & =  
\begin{vmatrix}
x-2 & 3x-6 & 2-x \\
2 & -3x & x-3\\
-3 & 2x & x+2 
\end{vmatrix}
\\
& = (x-2)
\begin{vmatrix}
1 & 3 & -1 \\
2 & -3x & x-3\\
-3 & 2x & x+2 
\end{vmatrix}
\\
& = (x-2)
\begin{vmatrix}
1 & 0 & 0 \\
2 & -3x-6 & x-1\\
-3 & 2x+9 & x-1 
\end{vmatrix}
\\
& = (x-2)(x-1)
\begin{vmatrix}
1 & 0 & 0 \\
2 & -3x-6 & 1\\
-3 & 2x+9 & 1 
\end{vmatrix}
\\
& = (x-2)(x-1)
\begin{vmatrix}
1 & 0 & 0 \\
5 & -5x-15 & 0\\
-3 & 2x+9 & 1 
\end{vmatrix}
\\
& = (x-2)(x-1)(-5x-15) \\
& = -5(x-2)(x-1)(x+3) \\
\end{align*}
$$
Therefore, the other roots are $1$ and $-3$.
A: $\begin{vmatrix}
x & -6 & -1 \\
2 & -3x & x-3\\
-3 & 2x & x+2 
\end{vmatrix} = 0$
$\begin{vmatrix}
x-2 & 3x-6 & 2-x\\
2 & -3x & x-3\\
-3 & 2x & x+2 
\end{vmatrix} = 0$
$R_1 \rightarrow R_1-R_2$
$(x-2)\begin{vmatrix}
1 & 3 & -1\\
2 & -3x & x-3\\
-3 & 2x & x+2 
\end{vmatrix} = 0$
$\begin{vmatrix}
1 & 3 & -1\\
0 & -3x-6 & x-1\\
0 & 2x+9 & x-1
\end{vmatrix} = 0$
$R_2 \rightarrow R_2-2R_1$,
$R_3 \rightarrow R_3+3R_1$
Now open the determinant using $C_1$
 clearly,one factor is (x-1). You get -3x-6= 2x+9, x=-3.
