Question

If one root of $\left|\begin{array}{ccc}7& 6& x\\ 2& x& 2\\ x& 3& 7\end{array}\right|=0$ is x =-9, find the other roots.

Answer 1

$\left|\begin{array}{ccc}7& 6& x\\ 2& x& 2\\ x& 3& 7\end{array}\right|=7(7x-6)-2(42-3x)+x(12-x^2)$
$=49x-42-84+6x+12x-x^3=-x^3+18x+49x-42-84=-x^3+67x-126$
(x+9) will be One of the factors of given cubic.

$-x^3+67x-126=(x+9)(-x^2+9x-14)=(-1)(x+9)(x^2-9x+14)=(-1)(x+9)(x^2-9x+14)=(-1)(x+9)(x^{-}7x-2x+14)=(-1)(x+9)(x-7)(x-2)=(x+9)(7-x)(x-2)=0\therefore x=-9,7,2$