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

2 Answers
Nov 7, 2016

Answer:

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)#

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Nov 30, 2016

Answer:

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]}