Solution:

Let the zeros of the given polynomial be α, β and γ. (3 zeros as it’s a cubic polynomial)

And given, the zeros are in A.P.

So, let’s consider the roots as

α = a – d, β = a and γ = a +d

Where a is the first term and d is a common difference.

From given f(x), a= 2, b= -15, c= 37 and d= 30

⇒ Sum of roots = α + β + γ = (a – d) + a + (a + d) = 3a = (-b/a) = -(-15/2) = 15/2

So, calculating for a, we get 3a = 15/2 ⇒ a = 5/2

⇒ Product of roots = (a – d) x (a) x (a + d) = a(a² –d²) = -d/a = -(30)/2 = 15

⇒ a(a² –d²) = 15

Substituting the value of a, we get

⇒ (5/2)[(5/2)² –d²] = 15

⇒ 5[(25/4) –d²] = 30

⇒ (25/4) – d² = 6

⇒ 25 – 4d² = 24

⇒ 1 = 4d²

∴ d = 1/2 or -1/2

Taking d = 1/2 and a = 5/2

We get,

The zeros as 2, 5/2 and 3

Taking d = -1/2 and a = 5/2

We get,

The zeros as 3, 5/2 and 2