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

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

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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:

comes*
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 http://www.youtube.com/watch?v=9Z1uNz__2xA
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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