"Expressing the roots of y5+11y4−77y3+132y2−77y+11=0 terms of of..." was answered by our experts

Sanai Huerta

Answered question

2022-04-04

Expressing the roots of

y^{5} + 11 y^{4} - 77 y^{3} + 132 y^{2} - 77 y + 11 = 0

terms of of

ζ_{11}

paganizaxpo3

Beginner2022-04-05Added 8 answers

t^{7} = 1

and

t \neq 1

, then

x = - 1 + \frac{2}{t + t^{- 1}}

is a root of

x^{3} + 7 x^{2} + 7 x - 7 = 0

but the six primitive 7th roots come in conjugate pairs giving the three roots of the cubic in x.
If

t^{11} = 1

and

t \neq 1

, then

z = t + t^{- 1} - 2

is a root of

z^{5} + 11 z^{4} + 44 z^{3} + 77 z^{2} + 55 z + 11 = 0

but the ten primitive 11th roots come in conjugate pairs giving the five roots of the quintic in z and the five roots of

y^{5} + 11 y^{4} - 77 y^{3} + 132 y^{2} - 77 y + 11 = 0

are from

y = (2 - u) (1 - u + u^{3})

where

u = t + t^{- 1}

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