If $\begin{vmatrix} x-4 & 2x & 2x\\2x& x-4 & 2x\\2x& 2x& x-4\end{vmatrix}= (A+Bx)(x-A)^2$, then the ordered pair (A, B) is : (where A and B are real numbers )

(-4, 5)

The correct answer is Option (1) → (-4, 5)

$\begin{vmatrix} x-4 & 2x & 2x\\2x& x-4 & 2x\\2x& 2x& x-4\end{vmatrix}$

$C_1→C_1+C_2+C_3$

$\begin{vmatrix} 5x-4 & 2x & 2x\\5x-4& x-4 & 2x\\5x-4& 2x& x-4\end{vmatrix}$

$⇒(5x-4)\begin{vmatrix} 1 & 2x & 2x\\1& x-4 & 2x\\1& 2x& x-4\end{vmatrix}$

$⇒R_2→R_2-R_1,R_3→R_3-R_1$

$(5x-4)\begin{vmatrix} 0 & 2x & 2x\\0& -x-4 & 0\\0& 0& -x-4\end{vmatrix}$

$⇒(x+4)^2(5x-4)=(A+Bx)(x-A)^2$

$A=-4,B=5$