pure math question

t = x + 1/x , so x^2 - tx + 1 = 0...★
Let x^5+9x^4-10x^3-10x^2+9x-1
= (x^2 - tx + 1)(x^3 + ax^2 + bx + 1)
= x^5+ (a - t)x^4 + (b- at + 1)x^3 + (a - bt + 1)x^2 + (b - t)x + 1
we have :
a - t = 9......(1)
b - at + 1 = -10......(2)
a - bt + 1 = -10......(3)
b - t = 9......(4)
From (1) and (4): a = b = t + 9 , sub to (2) :
t + 9 - t(t + 9) + 1 = -10
t^2 + 8t - 20 = 0
(t - 2)(t + 10) = 0
t = 2 or t = -10 , sub to ★ :
x^2 - 2x + 1 = 0 or x^2 + 10x + 1 = 0
x = 1 (double roots) or x = - 5 + 2√6 or -5 - 2√6
The 5 th root = -1 / [(1)(1)( -5 + 2√6)(-5 - 2√6)]
= -1 / (25 - 4*6)
= - 1

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