alpha, Beta, gamma are the zeroes of cubic polynomial x3 – 6×2 + p(x – 1) + 5. If alpha, beta, gamma are in A.P., then find the

September 9, 2021

By Gianna

1 thought on “alpha, Beta, gamma are the zeroes of cubic polynomial x3 – 6×2 + p(x – 1) + 5. If alpha, beta, gamma are in A.P., then find the<br”

1. Solution:

   Given –

   • f(x) = x³ – 6x² + p(x – 1) + 5
   • α, β and γ are the zeros of f(x).
   • Also, α, β and γ are in in A.P.

   We have to find out the product of Zeros.

   → f(x) = x³ – 6x² + px + (5 – p)

   As α,β and γ are in A.P.,

   → β – α = γ – β

   → 2β = α + γ – (i)

   We know that,

   → Sum of zeros of a cubic polynomial = -b/a

   Where,

   • b = Coefficient of x².
   • a = Coefficient of x³.

   Here,

   → a = 1

   → b = -6

   → c = p

   → d = 5 – p

   So,

   → Sum of roots = -b/a

   → Sum = 6

   So,

   → α + β + γ = 6

   From (i),

   → 3β = 6

   → β = 2

   Therefore,

   → f(β) = (2)³ – 6 × (2)² + p × (2 – 1) + 5

   → f(β) = 8 – 24 + p + 5

   → f(β) = p – 11

   As β is a zero,

   → f(β) = 0

   → p – 11 = 0

   → p = 11

   So,

   → d (Constant term in f(x)) = 5 – 11 = -6

   Therefore,

   → Product of zeros = -d/a

   = -(-6)/1

   = 6

   Answer:

   • Product of zeros (αβγ) = 6.

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