# Math Help - x^4+x^3+x^2+x+1 = 0

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

How to solve $x^4+x^3+x^2+x+1 = 0$

2. There are not any real solutions. A simple graph of the function will show this.

3. Originally Posted by Garas
How to solve $x^4+x^3+x^2+x+1 = 0$
Note that $\displaystyle 1+\cdots+x^4=\frac{1-x^5}{1-x}$ for $x\ne 1$ (but $x=1$ doesn't satisfy you're equation anyways). But, this is zero if and only if $1-x^5=0\implies x^5=1\implies x=1$ but we are determined that wasn't the case.

4. I know that there is no real solutions, but that equation have 4 complex solutions x&#94;4&#43;x&#94;3&#43;x&#94;2&#43;x&#43;1 &#61; 0 - Wolfram|Alpha

5. Originally Posted by Garas
I know that there is no real solutions, but that equation have 4 complex solutions x&#94;4&#43;x&#94;3&#43;x&#94;2&#43;x&#43;1 &#61; 0 - Wolfram|Alpha
Ok. Then use what I said, again. $x^5-1=0\implies x^5=1\implies x^5=e^{2\pi i}\implies x=e^{\frac{2\pi i}{5}},\cdots,e^{2\pi i}$

6. Originally Posted by Garas
I know that there is no real solutions, but that equation have 4 complex solutions x&#94;4&#43;x&#94;3&#43;x&#94;2&#43;x&#43;1 &#61; 0 - Wolfram|Alpha
Then you should have specified that you wanted complex solutions.

To follow what Drexel was saying,

$\displaystyle x^4 + x^3 + x^2 + x + 1 = \frac{1- x^5}{1 - x}$ for $\displaystyle x \neq 1$.

Therefore, if $\displaystyle x^4 + x^3 + x^2 + x + 1 = 0$

$\displaystyle \frac{1 - x^5}{1 - x} = 0$ for $\displaystyle x \neq 1$

$\displaystyle 1 - x^5 = 0$

$\displaystyle x^5 = 1$

$\displaystyle x^5 = e^{2\pi i}$

$\displaystyle x = e^{\frac{2\pi i}{5}}$.

There will be 5 solutions, all evenly spaced about a circle, so they will differ by an angle of $\displaystyle \frac{2\pi}{5}$.

So the solutions are

$\displaystyle \left\{e^{\frac{2\pi i}{5}}, e^{\frac{4\pi i}{5}}, e^{\frac{6\pi i}{5}}, e^{\frac{8\pi i}{5}}\right\}$ excluding $\displaystyle e^{\frac{10\pi i}{5}} = e^{2\pi i} = 1$.