The value of $4x^3-4x^2-7x+127\mathrm{when}x=\frac{4+5\sqrt{-1}}{2}is$

$ifx=\frac{4+5i}{2}$

$2x-4=5i$

${\left(2x-4\right)}^2=-25$

$4x^2-16x+41=0$
And $4x^3-4x^2-7x+127$ dividing
By $4x^2-16x+41$ by $4x^2-16x+41$

$4x^2-16x+41=\left(4x^2-16x+41\right)\left(x+3\right)+4$

 reminder is 4