Let `f(y) = 7y^(2)-(11)/(3)y -(2)/(3)`
`= 21 y^(2) - 11y - 2`
`= 21y^(2) - 14y +3y -2` [by splitting the middle term]
`=7y (3y -2)+1(3y -2)`
`=(3y-2) (7y+1)`
So, the value of `7y^(2)-(11)/(3)y -(2)/(3)` is zero when `3y - 2 = 0` or `7y +1 = 0`.
i.e., when `y = (2)/(3)` or `y =- (1)/(7)`.
So, the zeroes of `7y^(2) -(11)/(3)y - (2)/(3)` are `(2)/(3)` and `-(1)/(7)`
`:.` Sum of zeroes `= (2)/(3)-(1)/(7) =(14-3)/(21) =(11)/(21) = -((-11)/(3xx7))`
`=(-1) (("Coefficient of y"))/(("Coefficient of" y^(2)))`
and product of zeroes `= ((2)/(3)) (-(1)/(7)) =(-2)/(21) = (-2)/(3xx7)`
`= (-1)^(2) (("Constant term")/("Coefficient of " y^(2)))`
Hence, verified the relations between the zeroes and the coefficients of the polynomial.
