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in Polynomials by (32.2k points)
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If α, β, γ are the zeroes of the polynomial p(x) = 6x3 + 3x2 – 5x + 1, find the value of

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

1 Answer

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Best 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

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