Solving for $x$ in $x^5+x^4+x^3+x^2+x+1=0$

Question

So I was scrolling through the homepage of Youtube to see if there were any math equations that I thought that I might be able to solve when I came across this video by Blackpenredpen which was a video on solving equations but they get increasingly more awesome. For this question, I will be focusing on the "Awesome" level equation but here are all $5$:

Good:$\quad x+1=0\quad$Solution: $x=-1$

Great:$\quad x^2+x+1=0\quad$Solution: $x=\dfrac{1\pm\sqrt5}{2}$

Nice:$\quad x^3+x^2+x+1=0\quad$Solution: $x=-1$

Excellent:$\quad x^4+x^3+x^2+x+1=0\quad$Solution: $x=\dfrac{-\phi\pm\sqrt{\phi^2-4}}{2}$

Awesome:$\quad x^5+x^4+x^3+x^2+x+1=0\quad$Solution: We don't have this right now.

Here is my attempt at solving the "Awesome" level equation:

1. Try plugging in real numbers.
2. Attempt $1$: $1$
3. $1$ doesn't work because $6\neq0$
4. Attempt $2$: $-1$
5. $-1+1-1+1-1+1=0$

Therefore, the solution for $x$ in$$x^5+x^4+x^3+x^2+x+1=0$$is $-1$, and this will work for basically any problem like this with $x^n+x^{n-1}+x^{n-2}\dots+x^2+x+1=0$ where $n$ is an odd number, which is because they will all cancel out with each other because there are an even amount of terms.

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My question

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Is the solution that I have achieved correct, or what could I do to attain the correct solution/attain it more quickly?

Answer 1

Polynomials of the form $x^{2n+1}+ ...+x+1$ can be factored as $(x+1)(1+x^2+...+x^{2n})$. So for the case of $x^5+x^4+x^3+x^2+x+1$, we have that it is factorable as $(1+x)(1+x^2+x^4)$. Thus we can substitute $u=x^2$ into the quartic term, which results in $u^2+u+1$, which factors as $(u-\phi)(u+\frac{1}{\phi})$, so your roots of the quintic are $x=-1, \sqrt\phi, -\sqrt\phi, i\sqrt\frac{1}{\phi}, -i\sqrt\frac{1}{\phi}$.

Answer 2

If you only talk about real solution, then the answer is yes. Because

$$x^{2m+1}+x^{2m}+\dots+x+1=(x+1)(x^{2m}+x^{2m-2}+\dots+x^2+1)$$

Since

$$x^{2m}+x^{2m-2}+\dots+x^2+1\ge 1>0$$

So $x=-1$ is the only real root.

Answer 3

The solutions you're finding are correct for the odd powers, but :

• they aren't the only ones
• you're mistaken on the even powers

For clarification, $\frac{1 \pm \sqrt{5}}{2}$ are the solutions to $x^2-x-1=0$, which isn't what you wrote so I'd like to clear up confusion on that.

Also, that $\pm$ sign you'd have on the solutions for $x^2-x-1$ are hinting at the fact that there are more and more solutions every time you increase the polynomial's degree.

For any polynomial of the form $\displaystyle \sum_{k=0}^{n-1}{x^k}$, every element of the following set is a root of that polynomial : $$\left\lbrace e^{\frac{2i \pi k}{n}}, \quad 1 \leq k \leq n \right\rbrace$$

Some of these numbers are complex numbers, so for example, if you're only looking for the real solutions of $x^3+x^2+x+1$, yes, the solution is $-1$.

If you haven't seen the number $i$ yet, try doing some research on complex numbers, you're likely to have some fun.

Answer 4

Multiplying both sides of $$f_n(x) := x^n + x^{n - 1} + \cdots + x + 1 = 0$$ by $x - 1$ telescopes the sum, and rearranging gives $$x^{n + 1} = 1 .$$ So:

• If $n$ is odd, the only real solution is $-1$ (as $f_n(1) = n + 1 \neq 0$).
• If $n$ is even, there are no real solutions.

Over the complex numbers, the same argument shows that the solutions are exactly the $(n + 1)$st roots of unity other than $1$, namely, $$\exp \frac{k \pi i}{n + 1} = \cos \frac{k \pi}{n + 1} + i \sin \frac{k \pi}{n + 1}, \qquad k=1,\ldots,n .$$

Example The roots of $f_5$ are $-1$ and $\pm \frac{1}{2} \pm' \frac{\sqrt3}{2} i$