Finding roots of the polynomial $x^4+x^3+x^2+x+1$

Question

In general, how could one find the roots of a polynomial like $x^4+x^3+x^2+x^1+1$? I need to find the complex roots of this polynomial and show that $\mathbb{Q (\omega)}$ is its splitting field, but I have no idea of how to proceed in this question. Thanks in advance.

Answer 1

Since $(x-1)(x^4+x^3+x^2+x+1)=x^5-1$, the roots are the fifth roots of $1$, excluding $1$. Note that the set of fifth roots of $1$ is a group of prime order, so it is cyclic and any element is a generator. Thus, if $\omega$ is any of the roots of the polynomials, the full set of roots is given by $\omega,\omega^2,\omega^3,\omega^4$. In particular all roots belong to $\mathbb{Q}[\omega]$.

The roots are $$ \cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}\\ \cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5}\\ \cos\frac{6\pi}{5}+i\sin\frac{6\pi}{5}\\ \cos\frac{8\pi}{5}+i\sin\frac{8\pi}{5} $$

However, you can also determine them “more explicitly”. Rewrite the equation as $$ x^2+\frac{1}{x^2}+x+\frac{1}{x}+1=0 $$ and recall that $$ x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2 $$ so, setting $t=x+\dfrac{1}{x}$, the equation is $$ t^2+t-1=0 $$ Find the two roots $t_1$ and $t_2$; next solve the equations $$ x+\dfrac{1}{x}=t_1,\qquad x+\dfrac{1}{x}=t_2 $$

Answer 2

From the identity: $\; x^5-1=(x-1)(x^4+x^3+x^2+x++1)$, they are the complex $5$-th roots of unity: $$\mathrm e^{\tfrac{2\mathrm i k\pi}{5}},\quad k=1,2,3,4.$$

Answer 3

Generally, you may use the fact that the only irreducible polynomials are those of first and second degree with negative discriminant. So if you have a polynomial of higher degree, equating it to a product of quadratic and linear functions, finding the roots of these will amount to finding the roots of the original polynomial. Then again, when you go beyond fourth degree, you need to be lucky in order to be able to express the roots by radicals, but this one here is not such a bad case. Setting $$\begin{align} x^4+x^3+x^2+x+1&=(x^2+ax+1)(x^2+bx+1) \\ x^4+x^3+x^2+x+1&=x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1\end{align}$$ we see that $a+b=1$ and $ab=-1$. Solve for $a$ and $b$ and you'll be free to use the dear old quadratic formula!

Answer 4

One remark (which maybe you find useful): there is also slightly more general method for handling equations of the form $ax^4+bx^3+cx^2+bx+a=0$: divide both side by $x^2$ to obtain $ax^2+\frac{a}{x^2}+bx+\frac{b}{x}+c=0$ and then substitute $u=x+\frac{1}{x}$.