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Find the zeroes of the polynomial $p(x) = 3x^2 + 7x - 20$ and verify the relationship between its zeroes and the coefficients.
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(b) $p(x) = 3x^2 + 7x - 20 = 3x^2 + 12x - 5x - 20 = 3x(x+4) - 5(x+4) = (3x-5)(x+4)$ (1 Mark)
Zeroes are $-4, \frac{5}{3}$ (1 Mark)
Sum of the zeroes = $-4 + \frac{5}{3} = \frac{-12+5}{3} = -\frac{7}{3} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (1 Mark)
Product of zeroes = $-4 \times \frac{5}{3} = -\frac{20}{3} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ (1 Mark)
(b) $p(x) = 3x^2 + 7x - 20 = 3x^2 + 12x - 5x - 20 = 3x(x+4) - 5(x+4) = (3x-5)(x+4)$ (1 Mark)
Zeroes are $-4, \frac{5}{3}$ (1 Mark)
Sum of the zeroes = $-4 + \frac{5}{3} = \frac{-12+5}{3} = -\frac{7}{3} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (1 Mark)
Product of zeroes = $-4 \times \frac{5}{3} = -\frac{20}{3} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ (1 Mark)