2011/064) Solve 6x^5-41x^4+97x^3-97x^2 + 41x – 6 = 0

This as the coefficients are afternates signs and symmetric so they can be combined as below

6(x^5-1) – 41(x^4- x) + 97(x^3-1)
= 6(x-1)(x^4 + x^3 + x^2 + x + 1) – 41x(x-1)(x^2+x + 1) + 97x^2(x-1)
= (x-1)( 6x^4 + 6x^3 + 6x ^2 + 6x + 1 – 41x^3- 41 x^2 – 41 + 97 x^2 )
= (x-1)(6x^4 – 35x^3 + 62x^3 – 35 + 6)
So x= 1 is a solution and other factor is
(6x^4 – 35x^3 + 62x^2 – 35 + 6)
As the coefficients as symmetric we have combining similarly

6(x^4+1) – 35x(x^2+1) + 62 x^2 ...1
Taking x^2 out we get symmetric

X^2(6(x^2+ 1/x^2) - 35(x+1/x) + 62)

Put x+ 1/x = t o get
X^2(6(t^2-2) - 35 t + 62) = 0 and as x is not zero we get
(6(t^2-2) - 35 t + 62 = 0
Or 6t^2 - 35 t + 50 = 0
Ot (2t-5) (3t – 10) = 0
t= 5/2 or 10/3
now x + 1/x = 5/2 gives x ^2 + 1- 5/2 x= 0 or (4x^2 + 4x- 5) = 0 or (2x-1)(x-2) or x = ½ or 2
x + 1/x = 10/3 gives x ^2 + 1- 10/3 x= 0 or (3x^2 + 3x- 10) = 0 or (3x-1)(x-3) or x = 1/3 or 3

so solution = ½ , 1/3 , 1 , 2 , 3

you can also proceed from 1 putting x^2+1 = tx