Α,β and  γ are zeros of cubic polynomial and are in AP.

So, Let β=a ; α=a−d & γ=a+d

Polynomial=x3−12x2+44x+c

Sum of roots=1−(−12)​=12

So,a−d+a+a+d=12

3a=12a=4

Sum of products of two consecutive roots=44.

a(a−d)+a(a+d)+(a−d)(a+d)=44a2−ad+a2+ad+a2−d2=443a2−d2=443(4)2−d2=44d2=48−44=4d=±2

So, α=a−d=4+24−2​=62​

β=4β=a+d=4−24+2​=26​

So,Product (−c)=2×4×6

=−48

Solution:

Given, α, β, γ are the zeroes of the cubic polynomial x3 – 12x2 + 44x – c.

Let's assume that α, β, γ are in arithmetic progression (AP) such that β is the middle term.

So, β = (α + γ)/2

Sum of zeroes of a cubic polynomial is given by the formula:
α + β + γ = 12
Substituting β = (α + γ)/2, we get:
α + (α + γ)/2 + γ = 12
Simplifying the above equation, we get:
3α + 3γ = 24
α + γ = 8

Product of zeroes of a cubic polynomial is given by the formula:
αβγ = c
Substituting β = (α + γ)/2, we get:
α(α + γ)/2 × γ = c
Simplifying the above equation, we get:
α2γ + αγ2 = 2c

Now, we need to eliminate γ from the above two equations to get the value of α and then β. For this, we can use the formula α + γ = 8 and substitute γ as 8 - α.

α2(8 - α) + α(8 - α)2 = 2c
Simplifying the above equation, we get:
-2α3 + 36α - 128 = 2c

Since α, β, γ are zeroes of the cubic polynomial x3 – 12x2 + 44x – c, we can substitute α, β, and γ as the roots of the cubic equation. Hence, substituting β = (α + γ)/2, we get:

x3 – 12x2 + 44x – c = (x – α)(x – β)(x – γ)
x3 – 12x2 + 44x – c = (x – α)(x – (α + γ)/2)(x – (8 – α))

Expanding the above equation, we get:
x3 – (8α + 4γ)x2 + (α2 + 4αγ + γ2 - 8α - 32)x + c = 0

Comparing the coefficients of the above equation with the given cubic polynomial, we get:
-8α - 4γ = -12
α2 + 4αγ + γ2 - 8α - 32 = 44

Solving the above two equations simultaneously, we get:
α = 2
γ = 6
β = 4

Substituting the values of α and γ in the equation α2γ + αγ2 = 2c, we get:
2² × 6 + 2 × 6² = 2c
48 + 72 = 2c
c = 60

Hence, the value of c is 60.