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Find the zeroes of the polynomial $9s^2 - 6s + 1$ and verify the relationship between the zeroes and the coefficients of the given polynomial.
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$9s^2 - 6s + 1 = (3s - 1)(3s - 1)$. Zeroes are $\frac{1}{3}$ and $\frac{1}{3}$. Sum of zeroes $= \frac{1}{3} + \frac{1}{3} = \frac{2}{3} = \frac{-(-6)}{9} = \frac{-\text{Coefficient of } s}{\text{Coefficient of } s^2}$. Product of zeroes $= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} = \frac{\text{Constant term}}{\text{Coefficient of } s^2}$. ($1 + 1 + 1$ marks)