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Find the zeroes of the polynomial $q(x) = 8x^2 - 2x - 3$. Hence, find a polynomial whose zeroes are $2$ less than the zeroes of $q(x)$.
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$p(x) = 8x^2 - 2x - 3$. Zeroes are $-\frac{1}{2}$ and $\frac{3}{4}$. New zeroes are $-\frac{5}{2}$ and $-\frac{5}{4}$. Sum of new zeroes = $-\frac{5}{2} + (-\frac{5}{4}) = -\frac{15}{4}$. Product of new zeroes = $(-\frac{5}{2}) \times (-\frac{5}{4}) = \frac{25}{8}$. Required polynomial is $x^2 + \frac{15}{4}x + \frac{25}{8}$ or $8x^2 + 30x + 25$.