2x^4+5x^3+7x^2-x+41, when x=-2-3i - Vidyadip

Question

$2\text{x}^4+5\text{x}^3+7\text{x}^2-\text{x}+41,$ when $\text{x}=-2-\sqrt{3}\text{i}$

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Answer

$\text{x}=-2-\sqrt{3}\text{i}$ $\text{x}^2=(-2-\sqrt{3}\text{i})^2=4+4\sqrt{3}\text{i}+3\text{i}^2=1+4\sqrt{3}\text{i}$ $\text{x}^3=(1+4\sqrt{3}\text{i})(-2-\sqrt{3}\text{i})=-2-8\sqrt{3}\text{i}-\sqrt{3}\text{i}-12\text{i}^2=10+9\sqrt{3}\text{i}$ $\text{x}^4=(1-4\sqrt{3}\text{i})^2=1+8\sqrt{3}\text{i}+48\text{i}^2=-47+8\sqrt{3}\text{i}$ $2\text{x}^4+5\text{x}^3+7\text{x}^2-\text{x}+41$ $=2(-47)+8\sqrt{3}\text{i})+5(10-9\sqrt{3}\text{i})+7(1+4\sqrt{3}\text{i})-(-2-\sqrt{3}\text{i})+41$ $=-94+16\sqrt{3}\text{i}+50-45\sqrt{3}\text{i}+7+28\sqrt{3}\text{i}+2+\sqrt{3}\text{i}+41$ $=(-94+50+7+2+41)+(16\sqrt{3}\text{i}-45\sqrt{3}\text{i}+28\sqrt{3}\text{i}+\sqrt{3}\text{i})$ $=6+0$ $=6$

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