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 equal to:

We have to evaluate the determinant

$$\Delta \;=\;\begin{vmatrix} x-4 & 2x & 2x \\[2pt] 2x & x-4 & 2x \\[2pt] 2x & 2x & x-4 \end{vmatrix}$$

and express it in the factorised form $$(A+Bx)(x-A)^2.$$

First, we simplify the determinant with elementary row operations that do not change its value (they only make calculation easier).

Subtract the first row from the second and the third rows:

$$R_2 \to R_2-R_1,\; R_3 \to R_3-R_1.$$

This gives

$$ \Delta =\begin{vmatrix} x-4 & 2x & 2x\\[2pt] 2x-(x-4) & (x-4)-2x & 2x-2x\\[2pt] 2x-(x-4) & 2x-2x & (x-4)-2x \end{vmatrix} =\begin{vmatrix} x-4 & 2x & 2x\\[2pt] x+4 & -x-4 & 0\\[2pt] x+4 & 0 & -x-4 \end{vmatrix}. $$

Next, observe that both the second and third rows contain a common factor $$(x+4)$$. We factor this out:

$$ \Delta=(x+4)^2 \begin{vmatrix} x-4 & 2x & 2x\\[2pt] 1 & -1 & 0\\[2pt] 1 & 0 & -1 \end{vmatrix}. $$

Now we expand the remaining $$3\times3$$ determinant along the first row. Using the expansion formula $$ \begin{vmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{vmatrix} =a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31}), $$ we substitute

$$a_{11}=x-4,\;a_{12}=2x,\;a_{13}=2x,\; a_{21}=1,\;a_{22}=-1,\;a_{23}=0,\; a_{31}=1,\;a_{32}=0,\;a_{33}=-1.$$

Calculating each term one by one:

First term: $$(x-4)\bigl((-1)(-1)-0\cdot0\bigr)=(x-4)\cdot1=x-4.$$

Second term: $$-\,2x\bigl(1\cdot(-1)-0\cdot1\bigr)=-2x(-1)=+2x.$$

Third term: $$+\,2x\bigl(1\cdot0-(-1)\cdot1\bigr)=2x(0+1)=2x.$$

Adding them, we get

$$x-4+2x+2x=5x-4.$$

Therefore

$$\Delta=(x+4)^2(5x-4).$$

We are told that this must match $$(A+Bx)(x-A)^2.$$ Comparing the two factored forms term by term, we notice

$$x-A=x+4\;\Longrightarrow\;A=-4,$$

and then

$$A+Bx=-4+Bx=5x-4\;\Longrightarrow\;B=5.$$

Thus the ordered pair is $$(A,B)=(-4,5).$$

Hence, the correct answer is Option D.