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


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