Given that α, β, γ are the roots of x³ + 2x² + 3x + 4 = 0

Compare with x³ + bx² + cx + d = 0

b = 2, c = 3, d = 4

α + β + γ = – 6 = – 2

αβ + βγ + γα = c = 3

αβγ = – d = – 4

Given roots are 2α, 2β, 2γ

2α + 2β + 2γ = 2(α + β + γ)

= 2(– 2)

= – 4

(2α)(2β) + (2β)(2γ) + (2γ)(2α) = (4αβ + 4βγ + 4γα)

= 4(αβ + βγ + γα)

= 4(3)

= 12

(2α)(2β)(2γ) = 8(αβγ)

= 8(– 4)

= – 32

The equation is

x³ – x² (2α + 2β + 2γ) + x(4αβ + 4βγ + 4γα) – 8(αβγ)

= 0

⇒ x³ – x² (– 4) + x(12) – (– 32) = 0

⇒ x³ + 4x² + 12x + 32 = 0