Equation with one variable unknow how to solve easily I have 
$$\left|\begin{array}{ccc}
x & -2 & 3x-6 \\
2x & \phantom{-}0 & 2-x \\
-x & \phantom{-}5 & x-2 
\end{array}\right| = 0$$
How can I solve this with a fast way? I thought I can do 
$$x-2+3x-6=0 \implies 4x=-8 \implies x=-2$$ 
and I will continue with the other two, but I don't know if I am right. 
 A: Elements of the first column are all multiples of $x$; elements in the third column are all multiples of $x-2$. You can factor the determinant thusly:
$$0 = \left|\begin{array}{ccc}
\phantom{-}x & -2 & 3x-6 \\
2x & \phantom{-}0 & 2-x \\
-x & \phantom{-}5 & x-2 
\end{array}\right| = 
x(x-2) \left|\begin{array}{ccc}
\phantom{-}1 & -2 & \phantom{-}3 \\
\phantom{-}2& \phantom{-}0 & -1 \\
-1& \phantom{-}5 & \phantom{-}1 
\end{array}\right| =
37x(x-2) \quad\implies\quad x=0,2$$

To add a bit of detail ...
Determinant rules allow you to factor-out common multipliers from an individual colunmn (or row):
$$\left|\begin{array}{ccc}
pa & b & c \\
pd & e & f \\
ph & i & j
\end{array}\right| \;=\; p\left|\begin{array}{ccc}
a & b & c \\
d & e & f \\
h & i & j
\end{array}\right|$$
The justification is immediately clear if you're familiar with the relationship between a determinant and the volume of the parallelepiped determined by the column vectors. But it also makes sense if you pay attention to how determinants are expanded. Here's a quick example with a $2\times 2$:
$$\left|\begin{array}{cc} pa & b \\ pc & d \end{array}\right| = pa\cdot d -pc\cdot b = p(ad-cb) = p\left|\begin{array}{cc} a & b \\ c & d \end{array}\right|$$ 
Every term in the expansion includes the multiplier, which can be factored-out.
Anyway ... For the problem in question, I factored-out $x$ and $x-2$ from the first and third columns. Then I calculated the value of the remaining determinant (just to make sure it wasn't zero, in which case $x$ could be anything). 
A: This is wrong: 
$$x-2+3x-6=0 \implies 4x=-8 \implies x=-2 $$

This is the right way to do it 
$$\left|\begin{array}{ccc}
x & -2 & 3x-6 \\
2x & \phantom{-}0 & 2-x \\
-x & \phantom{-}5 & x-2 
\end{array}\right| = 0$$
Apply $R3 \rightarrow R3+R1$
$$\left|\begin{array}{ccc}
x & -2 & 3x-6 \\
2x & \phantom{-}0 & 2-x \\
0 & \phantom{-}3 & 4x-8 
\end{array}\right| = 0$$
Apply $R2\rightarrow R2-2R1$
$$\left|\begin{array}{ccc}
x & -2 & 3x-6 \\
0 & \phantom{-}4 & 14-7x \\
0 & \phantom{-}3 & 4x-8 
\end{array}\right| = 0$$
$$x\left(  4(4x-8)-3(14-7x)\right)=0$$
$$37x(x-2)=0 \implies x=0,2$$
A: You can also use the Laplace formula on the second column and get
$$\begin{vmatrix}
x & -2 & 3x-6 \\
2x & 0 & 2-x \\
-x & 5 & x-2 
\end{vmatrix} = 2 \begin{vmatrix} 2x & 2-x \\ -x & x-2 \end{vmatrix} - 5 \begin{vmatrix} x & 3x-6 \\ 2x & 2-x \end{vmatrix}=37x(x-2) $$
Hence $x \in \{0,2\}$.
