#"First we separate 4 x :"#
#= 4 x (2 x^4 + 7 x^3 + 3)#
#"Then we search the zeros of the remaining quartic equation."#
#2 x^4 + 7 x^3 + 3 = 0#
#"Substitute "x = 1/y#
#=> 2/y^4 + 7/y^3 + 3 = 0#
#=> 3 y^4 + 7 y + 2 = 0#
#=> y^4 + (7/3) y + 2/3 = 0#
#"Now we can apply Descartes method for solving a quartic"#
#"equation easily as the first two coefficients are zero."#
#=> y^4 + (7/3) y + 2/3 = (y^2 + k y + m)(y^2 - k y + n)#
#=> m + n - k^2 = 0#
#=> k n - k m = 7/3#
#=> m n = 2/3#
#=> k^2 = m + n#
#=> n - m = (7/3) / k#
#=> 2 n = k^2 + (7/3)/k#
#=> 2 m = k^2 - (7/3)/k#
#=> 4 m n = 8/3 = k^4 - (49/9)/k^2#
#=> k^6 - (8/3) k^2 - 49/9 = 0#
#"Substitute "k^2 = z#
#=> z^3 - (8/3) z - 49/9 = 0#
#"Substitute "z = r u#
#=> r^3 u^3 - (8/3) r u - 49/9 = 0#
#=> u^3 - ((8/3)/r^2) u - 49/(9 r^3) = 0#
#"Choose r so that the coefficient of u is -3."#
#=> r = sqrt(8)/3#
#=> u^3 - 3 u - 49/(9*8*sqrt(8)/27) = 0#
#=> u^3 - 3 u - 3*49/(8*sqrt(8)) = 0#
#"Substitute "u = t + 1/t#
#=> t^3 + 1/t^3 - 3*49/(8*sqrt(8)) = 0#
#"Substitute "v = t^3#
#=> v^2 - (3*49/(8*sqrt(8))) v + 1 = 0#
#"discriminant = "(3*49)^2/(64*8) - 4 = 6.1810257^2#
#=> v = ((3*49/(8*sqrt(8))) pm 6.1810257)/2#
#= 6.3387846 " if we take the solution with +"#
#"Substituting the variables back, yields"#
#=> t = root3(6.3387846) = 1.850697045#
#=> u = 2.391034#
#=> z = 2.25428847#
#=> k = 1.50142881#
#=> n = 1.90418185#
#=> m = 0.3501066#
#=> y^4 + (7/3) y + 2/3 =#
#(y^2 + 1.50142881 y + 0.3501066)*(y^2 - 1.50142881 y + 1.90418185)#
#"So we have two quadratic equations remaining."#
#"The first has positive discriminant, hence two real roots, but"#
#"the second one has negative discriminant, hence a pair of"#
#"complex conjugate roots, so we factor only the first further."#
#"discriminant = "k^2 - 4 m = 0.9240465^2#
#=> y = (-1.50142881 pm 0.9240465)/2#
#=> y = -1.2127377 " or "-0.2886911#
#=> x = -3.46391 " or "-0.8245806#
#=> 8 x^5 + 28 x^4 + 12 x#
#= 8 x (x + 3.46391)(x + 0.8245806)(x^2 - 0.78849 x + 0.52516)#
