Question

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

Answer 1

One real root and two pairs of conjugate complex roots.

Explanation:

We have that

`5+x+x^2+x^3+x^4+x^5=(x^6-1)/(x-1)+4` then

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

`{(y=x^6),(y = 5-4x):}`

We know that `x^6` is an even function and also that its epigraph is a convex set so the intersection of `y = 5-4x` 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
`(x+r)((x+a_1)^2+b_1^2)((x+a_2)^2+b_2^2)`

Answer 2

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]}