Find the roots of the polynomial

Question

I am trying to find the below roots of the polynomial -

$x^4 - x^3 + x^2 -x + 1 = 0$

Below is what I tried -

=$x^3 (x -1) + x (x - 1) + 1 = 0$

=$(x^3-1)(x+1)=-1$

Am I approaching the right way? But I am unable to proceed after this - how can I solve this - will this equation result in complex roots?

Answer 1

Write $t=x+1/x$ and divide given equation with $x^2$, so:

$$ x^2-x+1-1/x+1/x^2=0 \implies t^2-t-1=0$$

so $$t_{1,2} = {1\pm \sqrt{5}\over 2}$$

1.case $$x+{1\over x} = {1+ \sqrt{5}\over 2}\implies 2x^2-x(1+\sqrt{5})+2=0$$ $$ x_{1,2} = {1+\sqrt{5}\pm i\sqrt{10-2\sqrt{5}}\over 2}$$

2.case $$x+{1\over x} = {1- \sqrt{5}\over 2}\implies 2x^2-x(1-\sqrt{5})+2=0$$ $$ x_{3,4} = {1-\sqrt{5}\pm i\sqrt{2\sqrt{5}+10}\over 2}$$

Answer 2

Observe that $p(-x) = x^4+x^3+x^2+x+1 = \frac{x^5-1}{x-1} $ for any $x \neq 1$.

So $p(x) = \frac{x^5+1}{x+1} $, thus its roots are the primitive $10-$roots of unity.

Answer 3

Hint: One solution is given by $$x=1/4\sqrt {5}+1/4+i/4\sqrt {2}\sqrt {5-\sqrt {5}}$$

Answer 4

$$x^4-x^3+x^2-x+1=\Phi_{10}(x) $$ hence the roots of the LHS are the primitive tenth roots of unity, $\exp\left(\frac{\pi k i}{5}\right)$ for $k\in\{1,3,7,9\}$.
They can be found through $\cos\frac{\pi}{5}=\frac{1}{4}(1+\sqrt{5})$ and $\sin\frac{\pi}{5}=\sqrt{\frac{5-\sqrt{5}}{8}}$.

Answer 5

Here you can actually use a result from convolution. If we multiply the polynomial with $x+1$ we see we get $$x^5+1 = 0$$

This polynomial has the famous "roots of unity" which are complex numbers on the unit circle. So the roots to the equation are all $4$ roots with equal spacing except for the $x=-1$ one.