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.

Explanation

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.