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Find the zeroes of the polynomial $p(x) = 9x^2 - 6x - 35$ and verify the relationship between zeroes and its coefficients.
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Solution: $p(x) = 9x^2 - 6x - 35 = (3x - 7) (3x + 5)$
Zeroes of $p(x)$ are $\frac{7}{3}$ and $\frac{-5}{3}$
Sum of zeroes $= \frac{7}{3} - \frac{5}{3} = \frac{2}{3} = \frac{-(-6)}{9} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes $= \frac{7}{3} \times \frac{-5}{3} = \frac{-35}{9} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
Zeroes of $p(x)$ are $\frac{7}{3}$ and $\frac{-5}{3}$
Sum of zeroes $= \frac{7}{3} - \frac{5}{3} = \frac{2}{3} = \frac{-(-6)}{9} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes $= \frac{7}{3} \times \frac{-5}{3} = \frac{-35}{9} = \frac{\text{constant term}}{\text{coefficient of } x^2}$