माना α = –2, β = -3 और γ = -1
तब α + β + γ = (-2) + (-3) + (-1) = -2 – 3 – 1 = -6
αβ + βγ + γα = (-2)(-3) + (-3)(-1) + (-1)(-2) = 6 + 3 + 2 = 11
तथा αβγ = (-2)(-3)(-1) = -6
अतः अभीष्ट बहुपद = x3 – (α + β + γ)x2 + (αβ + βγ + γα)x – α·β·γ
= x3 – (-6)x2 + 11x – (-6)
= x3 + 6x2 + 11x + 6
