How do you find the number of complex, real and rational roots of 5+x+x^2+x^3+x^4+x^5=0?

Nov 7, 2016

One real root and two pairs of conjugate complex roots.

Explanation:

We have that

$5 + x + {x}^{2} + {x}^{3} + {x}^{4} + {x}^{5} = \frac{{x}^{6} - 1}{x - 1} + 4$ then

${x}^{6} + 4 x - 5 = 0$

includes de roots of

$5 + x + {x}^{2} + {x}^{3} + {x}^{4} + {x}^{5} = 0$

The solution for the real roots of

${x}^{6} + 4 x - 5 = 0$ is equivalent to find the intersections of

$\left\{\begin{matrix}y = {x}^{6} \\ y = 5 - 4 x\end{matrix}\right.$

We know that ${x}^{6}$ is an even function and also that its epigraph is a convex set so the intersection of $y = 5 - 4 x$ with its boundary has two solutions associated to two real roots, one of them being $x = 1$. The other root is the only real root for $5 + x + {x}^{2} + {x}^{3} + {x}^{4} + {x}^{5} = 0$ so concluding this last polynomial has one real root and two pairs of conjugate complex roots. This is because the polynomial has real coefficients.

Finally the structure for $5 + x + {x}^{2} + {x}^{3} + {x}^{4} + {x}^{5} = 0$ is
$\left(x + r\right) \left({\left(x + {a}_{1}\right)}^{2} + {b}_{1}^{2}\right) \left({\left(x + {a}_{2}\right)}^{2} + {b}_{2}^{2}\right)$ Nov 30, 2016

My answer is just to locate the only real root that is negative and close to -1.5 and also the point near (0, 5)

Explanation:

The real root can be bracketed to the desired sd-accuracy by

iterative methods. The 2-sd approximation is $- 1.5$. The other

aspects are very clear, in the nice answer by Cesareo R

graph{5+x+x^2+x^3+x^4+x^5 [-23.11, 4.91, -7.01, 6.99]}