Mathematics - Application of Derivative Question with Solution | TestHub

Given: $\left|\begin{array}{ccc}2\mathrm{x}+1& 4& 8\\ 2& 2\mathrm{x}& 2\\ 7& 6& 2\mathrm{x}\end{array}\right|=0$ $\begin{array}{rl}\therefore (2x+1)\left(4x^2-12\right)-4& (4x-14)\\ & +8(12-14x)=0\end{array}$ $\begin{array}{rl}\Rightarrow 8x^3-24x+4x^2-& 12-16x\\ & +56+96-112x=0\end{array}$ $\Rightarrow 8x^3+4x^2-152x+140=0$ $(x+5)$ is a factor of above equation $\begin{array}{rl}\therefore 8x^3+40x^2-36x^2-180x& \\ & +28x+140=0\\ \Rightarrow 8x^2(x+5)-36x(x+5)& +28(x+5)=0\end{array}$ $\Rightarrow (x+5)\left(8x^2-36x+28\right)=0$ $\Rightarrow (x+5)4\left(2x^2-9x+7\right)=0$ $\Rightarrow 4(x+5)\left(2x^2-7x-2x+7\right)=0$ $\Rightarrow 4(x+5)[x(2x-7)-1(2x-7)]=0$ $\Rightarrow 4(x+5)(2x-7)(x-1)=0$ $\Rightarrow x=-5,3.5,1$