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Find the zeroes of the polynomial $p(x) = 3x^2 - 4x - 4$. Hence, write a polynomial whose each of the zeroes is 2 more than zeroes of $p(x)$.
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$p(x) = 3x^2 - 4x - 4$. Zeroes are $-\frac{2}{3}$ and 2 (1 mark). New zeroes are $\frac{4}{3}$ and 4 ($\frac{1}{2}$ mark). Sum of new zeroes = $\frac{4}{3} + 4 = \frac{16}{3}$ ($\frac{1}{2}$ mark). Product of new zeroes = $\frac{4}{3} \times 4 = \frac{16}{3}$ ($\frac{1}{2}$ mark). Required polynomial is $x^2 - \frac{16x}{3} + \frac{16}{3}$ or $3x^2 - 16x + 16$ ($\frac{1}{2}$ mark).