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Find the zeroes of the polynomial $p(x) = 6x^2 + 13x - 5$ and verify the relationship between its zeroes and the coefficients.
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Solution: $p(x) = 6x^2 + 13x - 5 = (3x - 1)(2x + 5)$
Zeroes of $p(x)$ are $\frac{1}{3}, -\frac{5}{2}$ [1 mark]
Sum of zeroes $= \frac{1}{3} - \frac{5}{2} = \frac{-13}{6} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2}$ [1 mark]
Product of zeroes $= \frac{1}{3} \times \frac{-5}{2} = \frac{-5}{6} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ [1 mark]
Zeroes of $p(x)$ are $\frac{1}{3}, -\frac{5}{2}$ [1 mark]
Sum of zeroes $= \frac{1}{3} - \frac{5}{2} = \frac{-13}{6} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2}$ [1 mark]
Product of zeroes $= \frac{1}{3} \times \frac{-5}{2} = \frac{-5}{6} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ [1 mark]