#"There is no easy factorization here."#
#"So all one can do is apply general methods for cubic equations."#
#"I will show you how to apply Vieta's substitution : "#
#=> x^3 + 4 x^2 + 2.5 x + 8 = 0 " (after dividing through 2)"#
#"Now substitute "x = y-4/3#
#=> y^3 - (17/6) y + 254/27 = 0#
#"Substitute "y = sqrt(17/18) z#
#=> z^3 - 3 z + 10.2495625 = 0#
#"Substitute "z = t+1/t#
#=> t^3 + 1/t^3 + 10.2495625 = 0#
#"Substituting "u = t^3", yields the quadratic equation :"#
#=> u^2 + 10.2495625 u + 1 = 0#
#"A root of this quadratic equation is "u=-0.09851197#
#"Substituting the variables back, yields :"#
#t = root(3)(u) = -0.4618451#
#z = -2.62707324#
#y = -2.55305628#
#x = -3.88638961#
#"The other roots are complex : "-0.05680519 pm 1.43361046 i#
#"(they can be found by dividing away "(x+3.88638961))#
