If $\begin{vmatrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{vmatrix} = (A+Bx)(x-A)^2$ then $A + 2B$ equals
Step-by-Step Solution
Key Concept: Recognize that the given determinant has a circulant structure (each row is a cyclic shift) which allows column/row operations to extract common factors. Adding all columns produces a factor of (5x-4), then row operations reveal the squared factor (x+4)² by exploiting the symmetric difference pattern.
Perform column operations $C_1 \to C_1 + C_2 + C_3$ on the given determinant to factor out $(5x-4)$. Then use $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ to simplify. The determinant becomes $(5x-4)(x+4)^2 = (A+Bx)(x-A)^2$, giving $A=-4, B=5$, and $A+2B = -4+10 = 6$.
Correct Answer: 6