If the zeroes of the cubic polynomial x3 - 6x2 + 10 are of the form a, a + b and a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.
Let P(x) = x3 - 6x2 + 10
And (a), (a + b) and (a + 2b) are the zeroes of P(x).
Sum of the zeroes = - (coefficient of x2) ÷ coefficient of x3
α + β + γ = - b/a
a + (a + b) + (a + 2b) = - (- 6)
= 3a + 3b = 6
= a + b = 2
⇒ a = 2 - b (1)
Product of all the zeroes = - (constant term) ÷ coefficient of x3
αβγ = - d/a
a (a + b)(a + 2b) = - 10
= (2 - b) (2) (2 + b) = - 10
= (2 - b) (2 + b) = - 5
= 4 - b2 = - 5
⇒ b2 = 9
⇒ b = 
3
When b = 3, a = 2 - 3 = - 1 (from (1))
⇒ a = - 1
When b = - 3,a = 2 - (- 3) = 5 (from(1))
⇒ a = 5
Case1: when a = - 1 and b = 3
The zeroes of the polynomial are:
a = - 1 a + b = - 1 + 3 = 2 a + 2b = - 1 + 2(3) = 5
⇒ - 1, 2 and 5 are the zeroes
Case2: when a = 5,b = - 3
The zeroes of the polynomial are:
a = 5 a + b = 5 - 3 = 2 a + 2b = 5 - 2(3) = - 1
⇒ - 1, 2 and 5 are the zeroes
By both the cases the zeroes of the polynomial is - 1, 2, 5