x^4 +x^3 +x^2 +x+1=0


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Question Number 173839 by Khalmohmmad last updated on 19/Jul/22

$x^{4} + x^{3} + x^{2} + x + 1 = 0$
Commented by mr W last updated on 19/Jul/22

$x_{k} = \cos \frac{2 k π}{5} + i \sin \frac{2 k π}{5} (k = 1, 2, 3, 4)$
	Answered by mahdipoor last updated on 19/Jul/22

	$(x^{4} + . . . + 1) (x - 1) = 0 \times (x - 1) \Rightarrow$ $x^{5} - 1 = 0 \Rightarrow x = 1$ $b u t x i s r o o t o f (x - 1)$ $\Rightarrow\Rightarrow w b i t o u t a n s w e r$
	Answered by Rasheed.Sindhi last updated on 19/Jul/22

	$x^{4} + x^{3} + x^{2} + x + 1 = 0$ $(x - 1) (x^{4} + x^{3} + x^{2} + x + 1) = 0$ $x^{5} - 1 = 0$ $x^{5} = 1$ $x i s f i f t h r o o t o f u n i t y$ $L e t δ i s o n e 5 t h r o o t o f u n i t y .$ $A l l t h e 5 t h r o o t s a r e$ $δ, δ^{2}, δ^{3}, δ^{4}, δ^{5} = 1$ $x = 1 i s r o o t o f x - 1 a n d i s n o t r o o t$ $o f t h e g i v e n e q u a t i o n .$ $∴ δ, δ^{2}, δ^{3}, δ^{4} a r e t h e r o o t s o f t h e g i v e n$ $e q u a t i o n .$
	Answered by BaliramKumar last updated on 20/Jul/22

	$x^{4} + x^{3} + x^{2} + x + 1 = (x^{2} + a x + 1) (x^{2} + b x + 1) = 0$ $x^{4} + x^{3} + x^{2} + x + 1 = x^{4} + (a + b) x^{3} + (a b + 2) x^{2} + (a + b) x + 1 = 0$ $a + b = 1$ $a b + 2 = 1$ $a = \frac{\sqrt{5} + 1}{2} b = - \frac{\sqrt{5} - 1}{2}$ $x^{4} + x^{3} + x^{2} + x + 1 = [x^{2} + (\frac{\sqrt{5} + 1}{2}) x + 1] [x^{2} - (\frac{\sqrt{5} - 1}{2}) x + 1] = 0$ $x^{2} + (\frac{\sqrt{5} + 1}{2}) x + 1 = 0 & x^{2} - (\frac{\sqrt{5} - 1}{2}) x + 1 = 0$ $. . . . . . . . . . . . . . . . . . . . .$
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