Solve by determinant

Question

I have the following equation:

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

I tried solving this by solving for $0\times (det A)$ - $x\times (det b)$ + $1\times (det c)$ - $2\times (det d)$

and this got me to the equation $-x^{4} + 4x^{3} -2x^{2} -4x +3 = 0$ and I get the answer +-1 and 3 and this is correct, however I solved this by entering the equation online, Is there any easier way to solve for x in these types of equations? When I do determinants all the way I allmost allways seem to end up with complex equations that I cannot solve by my self.

Can I break out X from the matrix some way and simplify?

Answer 1

A reasonable attempt is to use the Rational root theorem, to find that if $p(x)$ has rational roots, then they must be in $\{\pm 1, \pm 3\}$.

Answer 2

Hint: A determinant is equal to zero if two of the rows (or columns) are linearly dependent. You can easily check that for $x=1$ the last two rows are linearly dependent, for example (they are the same). In general you can use Gussian elimination to simplify the problem.