If […] =(A+Bx)(x-A)^2, then the ordered pair (A,B) is equal to :

The correct answer is option (2) : $(-4, 5)$

We have,

$\begin{vmatrix}x-4 & 2x & 2x\\2x & x-4 & 2x\\2x & 2x & x-4\end{vmatrix}=(A+Bx)(x-A)^2$

Applying $C_1→C_1-C_3, C_2→C_2-C_3 $ on the LHS, we obtain

$\begin{vmatrix}-(x+4) & 0 & 2x\\0 & -(x+4) & 2x\\x+4 & x+4 & x-4\end{vmatrix}=(A+Bx)(x-A)^2$

$⇒(x+4)^2 \begin{vmatrix} -1 & 0 & 2x\\0&-1& 2x\\1&1&x-4\end{vmatrix}=(A+Bx) (x-A)^2$

Applying $R_3→R_3+R_1+R_2$ on the LHS, we obtain

$(x+4)^2 \begin{vmatrix} -1 & 0 & 2x\\0&-1& 2x\\0&0&5x-4\end{vmatrix}=(A+Bx) (x-A)^2$

$⇒(x+4)^2 (5x-4)=(A+Bx)(x-A)^2 $

$⇒A=-4, B= 5$