`4x^2 + 3x + 8`
`= 4x^2 + 3/2x + 3/2x + 9/16 + 119/16`
`= 2x.2x + 2x. 3/4 + 2x. 3/4 + 3/4. 3/4 + 119/16`
`= (2x.2x + 2x. 3/4) + (2x. 3/4 + 3/4. 3/4) + 119/16`
`= 2x. (2x + 3/4) + 3/4. (2x + 3/4) + 119/16`
`= (2x + 3/4).(2x + 3/4) + 119/16`
`= (2x + 3/4)^2 + 119/16`
Vì `(2x + 3/4)^2 >= 0 ∀ x ∈ R`
`=> (2x + 3/4)^2 + 119/16 >= 119/16 > 0 ∀ x ∈ R`
`=> 4x^2 + 3x + 8 > 0 ∀ x ∈ R`
`=>` Đa thức vô nghiệm