Given that zeroes of cubic polynomial `x^(3)-6x^(2)+3x +10` are of the form a,a+b,a+2b for some real numbers a and b, find the values of a and b as well as zeroes of the given polynomial.

Given that zeroes of cubic polynomial `x^(3)-6x^(2)+3x +10` are of the

1.	Given that zeroes of cubic polynomial `x^(3)-6x^(2)+3x +10` are of the form a,a+b,a+2b for some real numbers a and b, find the values of a and b as well as zeroes of the given polynomial.
Answer» Let `f(x) = x^(3) - 6x^(2) +3x +10` Given that, `a(a+b)` and `(a+2b)` are the zeroes of `f(x)`. The, Sum of the zeroes `=- (("Coefficient of" x^(2)))/(("Coefficient of" x^(3)))` `rArr a + (a+b) + (a+2b) =- ((-6))/(1)` `rArr 3a +3b = 6` `rArr a +b = 2` ...(i) Sum of product of two zeroes at a time `= (("Coefficient of x")/("Coefficient of" x^(3)))` `rArr a(a+b) +(a+b) (a+2b)+a(a+2b) = (3)/(1)` `rArr a (a+b) +(a+b) {(a+b)+b} +a{(a+b)+b} = 3` `rArr 2a +2 (2+b) +a (2+b) = 3` [using Eq.(i)] `rArr 2a +2 (2+2-a) + a (2+2-a) = 3` [using Eq(i)] `rArr 2a +8 - 2a +4a - a^(2) = 3` `rArr -a^(2) +8 = 3 - 4a` `rArr a^(2)-4a - 5 = 0` Using factorisation method, `a^(2) -5a +a - 5 = 0` `rArr a(a-5) +1(a-5) = 0` `rArr (a-5) (a+1) = 0` `rArr a =- 1,5` when `a =- 1`, then `b = 3` when `a = 5`, then `b =- 3` [using Eq.(i)] `:.` Required zeroes of `f(x)` are When `a=- 1` and `b = 3` then, `a (a+b), (a+2) =- 1, (-1+3), (-1+6)` or `-1,2,5` When `a = 5` and `b =- 3` then `a, (a+b), (a+2b) = 5, (5-3), (5-6)` or `5,2,-1`. Hence, the required valus of a and b are `a =- 1` and `b = 3` or `a = 5, b =- 3` and the zeroes are `- 1,2` and 5.