2009/013) Solve x^4+3x^3-8x^2+3x+1=0 for x

Generally 4th order equation is not easy to solve but this is a special case as this is symetrical that is coefficent of x^4 is same as x^0( and of x^3 same as of x)

so we devide by x^2 to get

x^2+3x-8+3/x+1/x^2 =0
or (x^2+1/x^2) + 3(x+1/x) - 8 = 0

put x+1/x = t to get

t^2-2 + 3t - 8 =0
or t^3+3y-10 = 0
(t-2)(t+5) = 0

t- 2 = 0 =>
x+1/x-2 = 0
or x^2+1-2x = 0
(x-1)^2 = 0

or x = 1 ( a double root)
2nd part
t+-5 = 0
=>x^2+5x + 1 = 0
x= (-5 +/-sqrt(21)/2 using (-b+/-sqrt(b^2-4ac))/2

= (-5 + sqrt(21))/2, (-5 - sqrt(21))/2,

so 4 roots are (-5 + sqrt(21))/2, (-5 - sqrt(21))/2, 1(Double root)