#"First we search the zeros"#
#x^5 + 3 x^2 - x = x(x^4 + 3 x - 1)#
#x^4 + 3 x - 1 = (x^2 + a x + b)(x^2 - a x + c)#
#=> b+c-a^2 = 0 ,#
#" "a(c-b) = 3 ,#
#" "bc = -1#
#=> b+c = a^2, " "c-b = 3/a#
#=> 2c = a^2+3/a, " "2b = a^2-3/a#
#=> 4bc = a^4 - 9/a^2 = -4#
#"Name k = a²"#
#"Then we get the following cubic equation"#
#k^3 + 4 k - 9 = 0#
#"Substitute k = r p :"#
#r^3 p^3 + 4 r p - 9 = 0#
#=> p^3 + (4/r^2) p - 9/r^3 = 0#
#"Choose r so that 4/r² = 3 => r = "2/sqrt(3)#
#"Then we get"#
#=> p^3 + 3 p - (27/8) sqrt(3) = 0#
#"Substitute p = t - 1/t :"#
#=> t^3 - 1/t^3 - (27/8) sqrt(3) = 0#
#=> t^6 - (27/8) sqrt(3) t^3 - 1 = 0#
#"Substitute u = t³, then we get a quadratic equation"#
#=> u^2 - (27/8) sqrt(3) u - 1 = 0#
#disc : 3*(27/8)^2 + 4 = 2443/64#
#=> u = ((27/8) sqrt(3) pm sqrt(2443)/8)/2#
#=> u = (27 sqrt(3) pm sqrt(2443))/16#
#"Take the solution with the + sign : "#
#u = 6.0120053#
#=> t = 1.8183317#
#=> p = 1.2683771#
#=> k = 1.4645957#
#=> a = 1.2102048#
#=> b = -0.50716177#
#=> c = 1.9717575#
#x^4 + 3 x - 1 = (x^2 + a x + b)(x^2 - a x + c)#
#"So the roots are"#
#x = (-a pm sqrt(a^2-4*b))/2#
#=> x = -0.6051024 pm 0.93451094#
#=> x = -1.53961334 " OR " 0.32940854#
#"And"#
#x = (a pm sqrt(a^2-4*c))/2#
#=> x = "not real as " a^2-4*c < 0#
#"So we have three zeros for our original quintic equation :"#
#x = = -1.53961334 " OR " 0 " OR " 0.32940854#
#"The end behavior is"#
#lim_{x->-oo} = -oo" , and"#
#lim_{x->+oo} = +oo."#
#"So we have"#
#-oo "........." -1.53961334 "........." 0 ".........." 0.32940854 "........" +oo#
#------0++++0----0+++++#
