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Find the value of $\;2x^{4} + 5 x^{3} +7x^{2} - x+41\;,$ when $\;x=-2-\sqrt{3} i$

$(a)\;1\qquad(b)\;0\qquad(c)\;6\qquad(d)\;4$

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1 Answer

Answer : $\;6$
Explanation :
$\;x=-2-\sqrt{3} i$
$\;x+2=-\sqrt{3} i$ => $\;x^{2} +4x+7=0$
$\;2x^{4} + 5 x^{3} +7x^{2} - x+41\;= (x^{2}+4x+7)(2x^{2} -3x+5)+6$
$= 0 \times (2x^{2} -3x +5) +6 =6 $
answered Apr 17, 2014 by yamini.v
 

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