If α, β, γ are the zeroes of the polynomial p(x) = 6x3 + 3x2 – 5x

1.	If α, β, γ are the zeroes of the polynomial p(x) = 6x3 + 3x2 – 5x + 1, find the value of\((\frac{1}{\alpha}+\frac1{\beta}+\frac1{\gamma})\).
Answer» Given: p(x) = 6x³ + 3x² – 5x + 1 = 6x³ – (–3) x² + (–5) x – 1 Comparing the polynomial with x³ – x² (α + β + γ) + x (αβ + βγ + γα) – αβγ, we get: αβ + βγ + γα = –5 and αβγ = – 1 ∴ \((\frac{1}{\alpha}+\frac1{\beta}+\frac1{\gamma})\) = \((\frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma})\) = \((\frac{-5}{-1})\) = 5

If α, β, γ are the zeroes of the polynomial p(x) = 6x3 + 3x2 – 5x + 1, find the value of\((\frac{1}{\alpha}+\frac1{\beta}+\frac1{\gamma})\).

Answer»

Given:

p(x) = 6x³ + 3x² – 5x + 1

= 6x³ – (–3) x² + (–5) x – 1

Comparing the polynomial with x³ – x² (α + β + γ) + x (αβ + βγ + γα) – αβγ,

we get:

αβ + βγ + γα = –5

and αβγ = – 1

∴ \((\frac{1}{\alpha}+\frac1{\beta}+\frac1{\gamma})\)

= \((\frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma})\)

= \((\frac{-5}{-1})\)

= 5