How to solve this determinant equation in a simpler way 
Question Statement:-
  Solve the following equation
  $$\begin{vmatrix}
x & 2 & 3 \\ 
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=0$$


My Solution:-
$$\begin{vmatrix}
x & 2 & 3 \\ 
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=
\begin{vmatrix}
x+5 & 2 & 3 \\ 
x+5 & x & 1 \\
x+7 & 2 & 5 \\
\end{vmatrix} \tag{$C_1\rightarrow C_1+C_2+C_3$}$$
$$=\begin{vmatrix}
0 & 2 & 3 \\ 
0 & x & 1 \\
2 & 2 & 5 \\
\end{vmatrix}+
(x+5)\begin{vmatrix}
1 & 2 & 3 \\ 
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}\tag{1}$$
On opening the first determinant in the last step above we get $2(2-3x)$.
On simplifying the secind determinant we get,
$$(x+5)\begin{vmatrix}
1 & 2 & 3 \\ 
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}=(x+5)\begin{vmatrix}
1 & 2 & 3 \\ 
0 & x-2 & -2 \\
0 & 0 & 2 \\
\end{vmatrix} 
(R_2\rightarrow R_2-R_1) 
(R_3\rightarrow R_3-R_1)$$
$=2(x+5)(x-2)$
Substituting the values obtained above in $(1)$, we get
$$=\begin{vmatrix}
0 & 2 & 3 \\ 
0 & x & 1 \\
2 & 2 & 5 \\
\end{vmatrix}+
(x+5)\begin{vmatrix}
1 & 2 & 3 \\ 
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}=2(2-3x)+2(x+5)(x-2)=2(2-3x+x^2+3x-10)=2(x^2-8)$$
Now, as $\begin{vmatrix}
x & 2 & 3 \\ 
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=0$, $\therefore 2(x^2-8)=0\implies x=\pm2\sqrt2$
As you can see there was lot of work in my solution so if anyone can provide me with some techniques to solve it faster, or a technique which includes less amount of pen and more thinking.
 A: Notice that the first two columns are proportional, hence linearly dependent, if
$$\frac x 2 = \frac 4 x$$
which is the same as $x = \pm 2 \sqrt{2}$ after solving. Convince yourself that the third column can never be written as a linear combination of the first two by noticing that the first and second columns have equal components in rows 1 and 3; you can work this into rigorous proof that the above equation gives all solutions.
A: Subtract the first row from the last and use Laplace's formula for the third row: $$\begin{vmatrix}
x & 2 & 3 \\ 
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}= \begin{vmatrix}
x & 2 & 3 \\ 
4 & x & 1 \\
0 & 0 & 2 \\
\end{vmatrix} = 2(-1)^{3+3}\begin{vmatrix}
x & 2\\ 
4 & x 
\end{vmatrix} = 2(x^2-8) = 2(x-2\sqrt{2})(x+2\sqrt{2})=0.$$ Now the roots are $x = \pm 2\sqrt{2}$.
A: Just expand the determinant! We want
$$
5x^2+2x+24-2x-40-3x^2=0
$$
which simplifies to $2x^2-16=0$.
(In general, finding determinants via row/column relations is faster when a matrix is large. But $3 \times 3$ matrices aren't that large yet...)
