If \begin{vmatrix} x & 4 & 2x \ 2x & x & 2x \ 2x & 2x & x \end{vmatrix} = (A+Bx)(x-A)^2, then the ordered pair (A,B) is equal to (4,5) , (-4,-5) , (-4,3) , (-4,5)

\begin{vmatrix} x & 4 & 2x \ 2x & x & 2x \ 2x & 2x & x \end{vmatrix} = (A+Bx)(x-A)^2
Consider D = \begin{vmatrix} x & 4 & 2x \ 2x & x & 2x \ 2x & 2x & x \end{vmatrix}
Applying R1 \rightarrow R1+R2+R3
D = \begin{vmatrix} 5x & 5x & 5x \ 2x & x & 2x \ 2x & 2x & x \end{vmatrix}
D = (5x) \begin{vmatrix} 1 & 1 & 1 \ 2x & x & 2x \ 2x & 2x & x \end{vmatrix}
Applying R2 \rightarrow R2-R3
D = (5x) \begin{vmatrix} 1 & 1 & 1 \ 0 & -x & x \ 2x & 2x & x \end{vmatrix}
D = (5x)[1(-x(x)-x(2x)) - 1(0-x(2x)) + 1(0-(-x)(2x))]
D = (5x)[-x^2-2x^2+2x^2+2x^2]
D = (5x)[x^2+8x+16]
D = (5x)(x+4)^2
D = (-4+5x)(x-(-4))^2
Hence, A=-4, B=5