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## anonymous 4 years ago Solve x^4+x^3+x^2+x+1 = 0 by hand

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1. Mr.Math

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 $$x^5=1$$.

2. anonymous

Why is x^4+x^3+x^2+x+1 = (x^5-1)/(x-1)

3. Mr.Math

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.

4. Mr.Math

comes*

5. anonymous

o right

6. anonymous

but does that give you all of the roots?

7. Mr.Math

$$x^5-1$$ has all the roots of your problem and one more root which is $$x=1$$, so you have to exclude it.

8. Mr.Math

You know how to solve $$x^5=1$$?

9. anonymous

1 and 4 complex solutions

10. Mr.Math

Exactly. The four complex solutions are what you're looking for, as you can see from the factorization of $$x^5-1$$.

11. anonymous

im not really sure about how to solve x^5=1

12. Mr.Math

Have a look at this http://www.youtube.com/watch?v=9Z1uNz__2xA

13. anonymous

o ty

14. anonymous

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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