Respuesta :

discipulus

Answer:

[tex] 5{x}^{3} - 4x + 1 = 0\\ let \: x = - 1 \\ 5{( - 1)}^{3} - 4( - 1)+ 1 = 0 \\ - 5 + 4 + 1 = 0 \\ 0 = 0 \\ (x + 1)(5 {x}^{2} - 5x + 1) =0 \\ D = {b}^{2} - 4ac \\ D = {5}^{2} - 4( - 5)(1) = 25 + 20 = 45 \\ \therefore \: D > 0 \\ all \: roots\: are \: real[/tex]