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Exercise 1.2 · Q81

Q.Find the value of 2x4+5x3+7x2−x+412x^4+5x^3+7x^2-x+41, if x=−2−3 ix = -2-\sqrt3\,i.

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x=−2−sqrt3,ix=-2-\\sqrt3\\,i. x2=(−2−sqrt3i)2=4+4sqrt3i+3i2=4+4sqrt3i−3=1+4sqrt3ix^2=(-2-\\sqrt3i)^2=4+4\\sqrt3i+3i^2=4+4\\sqrt3i-3=1+4\\sqrt3i. x3=x2x=(1+4sqrt3i)(−2−sqrt3i)=−2−sqrt3i−8sqrt3i−4cdot3,i2=−2−9sqrt3i+12=10−9sqrt3ix^3=x^2x=(1+4\\sqrt3i)(-2-\\sqrt3i)=-2-\\sqrt3i-8\\sqrt3i-4\\cdot3\\,i^2=-2-9\\sqrt3i+12=10-9\\sqrt3i. x4=x3x=(10−9sqrt3i)(−2−sqrt3i)=−20−10sqrt3i+18sqrt3i+9cdot3,i2=−20+8sqrt3i−27=−47+8sqrt3ix^4=x^3x=(10-9\\sqrt3i)(-2-\\sqrt3i)=-20-10\\sqrt3i+18\\sqrt3i+9\\cdot3\\,i^2=-20+8\\sqrt3i-27=-47+8\\sqrt3i. Now $2x^4+5x^3+7x^2-x+41=2(-47+8\sqrt3i)+5(10-9\sqrt3i)+7(1+ …

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