Find the value of the determinant $\scriptsize \begin{vmatrix} 8-x & -10 & 6 \\ 8 & -9-x & -1 \\ 16 & -15 & 9-x \end{vmatrix}$ I need to determine the value of the determinant and have it represented as product of factors. 
\begin{vmatrix}
 8-x & -10 & 6 \\ 
 8 & -9-x & -1 \\ 
 16 & -15 & 9-x
\end{vmatrix}
Also I need to use the properties of determinants since using Sarrus' or Laplace's Rule right away won't do me much good in a sense that I would obtain a polynomial of a degree higher than 2 that needs to be factored, and I want to (and I need to) avoid that.
So far I've rewritten it as:
\begin{vmatrix}
 8-x & -10  &  6\\ 
 0 & \frac{8+x+x^2}{8-x} & \frac{-56+x}{8-x} \\ 
 0 & \frac{40+15x}{8-x} & \frac{-24-17x+x^2}{8-x} 
\end{vmatrix}
But when I multiply the 2nd row by $\frac{-(40+15x)}{8+x+x^2}$ and add it to 3rd row I do obtain a triangular determinant (value of which is obtained by multiplying the elements on the main diagonal) but it's still in the form of an unfactored polynomial. That requires then further polynomial factorisation which I'm not allowed. I need to have it factored right from the determinant.
Any smarter way to do so maybe? Other than this algorithmic Gauss approach (rewriting the determinant to its equivalent triangular form).
 A: I don't think what you want is possible at all - at least not in a general setting. In the process of triangulization you will almost surely always get non-factorized polynomials on the main diagonal (like in your example) which will eventually lead you to a non-factorized characteristic polynomial. 
Just imagine there would be a way, by simple gauss-elimination, to achieve an upper triangle matrix with factorized main diagonal. This would essentially mean, we would have found a way to factorized cubic polynomials (not by using Cardano's formula). I am not an expert in complexity nor in information theory, but it seems very unlikely that this could work.
Instead I would recommend to use either 


*

*Cardano's formula

*or the Rational root theorem (as mentioned in the comments already), after using Sarrus to calculate the determinant


While Cardano's formula works in the general case, we would have to calculate a lot and errors are likely. If we are dealing with an integer matrix (at least in this case) we can apply the rational root theorem safely (like in your example) since we'll always have
$$
 \phi(x)=\pm x^{n}+a_{n-1}x^{n-1}+\cdots +a_{0}
$$
with $a_i\in \mathbb Z$. So we can just test all the divisors of $a_0$ to find the roots of your polynomial (or find one and then jump to polynomial division). So in your example we really just have to test the divisors of ($\pm$ is relevant)
$$
a_0=256 \text{ which are } 1,2,\ldots,2^8
$$
which gives eventually the factors $x_{1,2}=-4$ and $x_3=16$ and finally
$$
\phi(x)=-(x - 16) (x + 4)^2
$$
A: The determinant is a cubic.  It is $-x^3+8x^2+112x+256$  The rational root theorem says the roots, if rational, are positive or negative factors of $256$, which doesn't give too many to try.  For class problems, the roots of polynomials higher than quadratics are usually integers, so those are the things to try.  This one factors as $-(x-16)(x+4)^2$
