If $x=-2-\sqrt{3} i$, where $i=\sqrt{-1}$, then the value of $2 x^4+5 x^3+7 x^2-x+41$ is

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Aicharya · Accepted Answer

$\begin{aligned} & x=-2+\sqrt{3} i \ & \Rightarrow x+2=-\sqrt{3} i \ & \Rightarrow(x+2)^2=(-\sqrt{3} i)^2 \ & \Rightarrow x^2+4 x+4=-3 \ & \Rightarrow x^2+4 x+7=0 \ & 	ext { Now } 2 x^4+5 x^3+7 x^2-x+41 \ & =\left(2 x^2-3 x+5ight)\left(x^2+4 x+7ight)+6 \ & =\left(2 x^2-3 x+5ight) 	imes 0+6 \ & =6\end{aligned}$

If $x=-2-\sqrt{3} i$, where $i=\sqrt{-1}$, then the value of $2 x^4+5 x^3+7 x^2-x+41$ is

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