Question

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

Answer 1

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 \(x^5=1\).

Answer 2

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

Answer 3

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.

Answer 4

comes*

Answer 5

o right

Answer 6

but does that give you all of the roots?

Answer 7

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

Answer 8

You know how to solve \(x^5=1\)?

Answer 9

1 and 4 complex solutions

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

o ty

Answer 14

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.