Question

If $\left|\begin{array}{ccc}x-4& 2x& 2x\\ 2x& x-4& 2x\\ 2x& 2x& x-4\end{array}\right|=(A+Bx){(x-A)}^2$ Then the ordered pair (A,B) is equal to

Answer 1

$\begin{array}{c}\left(x-4\right)\left[\left(x-4\right)\left(x-4\right)-4x^2\right]-2x\left[2x\left(x-4\right)-4x^2\right]\\ +2x\left[4x^2-2x\left(x-4\right)\right]=\left(A+Bx\right){\left(x-A\right)}^2\\ \left(x-4\right)\left[x^2-8x+16-4x^2\right]-2x\left[2x^2-8x-4x^2\right]\\ 2x\left[4x^2-2x^2+8x\right]=\left(A+Bx\right){\left(x-A\right)}^2\\ \left(x-4\right)\left[-3x^2-8x+16\right]-2x\left[-2x^2-8x\right]=\left(A+Bx\right){\left(x-A\right)}^2\\ +2x\left[2x^2+8x\right]\\ -3x^3-8x^2+16x+12x^2+32x-64+4x^3+16x^2=\left(A+Bx\right){\left(x-A\right)}^2+4x^3+16x^2\\ 5x^3+36x^2+48x-64=\left(A+Bx\right)\left(x^2-2Ax+A^2\right)\\ =A{x}^2-2A^{2}x+A^3+B{x}^3-2AB{x}^2+A^{2}Bx\\ Oncomparingwith\underset{–}{x^3.}\\ 5=B\\ orB=5\\ oncomparing\underset{–}{x}.\\ 48=A^{2}B-2A^2\\ 48=5A^2-2A^2\\ 48=3A^2\\ \frac{48}{3}=A^2\\ 16=A^2\\ A=4\\ \left(A,B\right)=\left(4,5\right)Ans.\end{array}$