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Solve x^4+x^3+x^2+x+1 = 0 by hand

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You're studying complex numbers I assume. I will give you a hint: \[x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1}.\] Solve it now by first finding the roots of unity of \(x^5=1\).
Why is x^4+x^3+x^2+x+1 = (x^5-1)/(x-1)
This come from the factorization \(x^5-1=(x-1)(x^4+x^3+x^2+x+1)\). If you don't already know that. Then you can easily notice that \(1\) is a root of \(x^5-1\), and then you can use synthetic division to get the above factorization.

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Other answers:

o right
but does that give you all of the roots?
\(x^5-1\) has all the roots of your problem and one more root which is \(x=1\), so you have to exclude it.
You know how to solve \(x^5=1\)?
1 and 4 complex solutions
Exactly. The four complex solutions are what you're looking for, as you can see from the factorization of \(x^5-1\).
im not really sure about how to solve x^5=1
Have a look at this
o ty
Or First divide the entire equation by x^2 \[x^2+x+1+\frac{1}{x}+\frac{1}{x^2}=0 \implies x^2+\frac{1}{x^2}+x+\frac{1}{x}+1=0\] Let u = x+1/x \[u^2 = x^2+2+\frac{1}{x^2}\] The original equation now becomes: \[u^2-2+u+1 = 0 \implies u^2+u-1=0\] Use quadratic equation then substitute u = x+1/x back and then solve the quadratic again.

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