If α,β and γ are roots of 4x3+8x2−x−2=0 , then the value of 4(α+1)(β+1)(γ+1)αβγ is equal to

4x3+8x2−x−2=4(x−α)(x−β)(x−γ)Put x=−1, we get 3=4(−1−α)(−1−β)(−1−γ)Or (α+1)(β+1)(γ+1)=−34Also αβγ=12 ∴4(α+1)(β+1)(γ+1)αβγ=−6

If α,β and γ are roots of 4x3+8x2−x−2=0 , then the value of 4(α+1)(β+1)(γ+1)αβγ is equal to

4x3+8x2−x−2=4(x−α)(x−β)(x−γ)Put x=−1, we get 3=4(−1−α)(−1−β)(−1−γ)Or (α+1)(β+1)(γ+1)=−34Also αβγ=12 ∴4(α+1)(β+1)(γ+1)αβγ=−6

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If $α, β$ and $γ$ are roots of $4 x^{3} + 8 x^{2} - x - 2 = 0$ , then the value of $\frac{4 (α + 1) (β + 1) (γ + 1)}{α β γ}$ is equal to

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$- 6$

$6$

$\frac{- 3}{2}$

$\frac{3}{2}$

answer is D.

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Detailed Solution

$4 x^{3} + 8 x^{2} - x - 2 = 4 (x - α) (x - β) (x - γ)$
Put $x = - 1,$ we get $3 = 4 (- 1 - α) (- 1 - β) (- 1 - γ)$
Or $(α + 1) (β + 1) (γ + 1) = \frac{- 3}{4}$
Also $α β γ = \frac{1}{2}$
$∴ \frac{4 (α + 1) (β + 1) (γ + 1)}{α β γ} = - 6$