Question

alpha, Beta, gamma are the zeroes of cubic polynomial x3 – 6x2 + p(x - 1) + 5. If alpha, beta, gamma are in A.P., then find the
product of all the zeroes.

I need steps also
Irrelevant answers will be reported!​

Answer 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.