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Find the zeroes of the polynomial $p(x) = 6x^2 - 5x - 1$. Hence, obtain a polynomial each of whose zeroes is three times the zeroes of $p(x)$.
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$p(x) = 6x^2 - 5x - 1 = (x - 1)(6x + 1)$
$\therefore$ Zeroes are $1, -\frac{1}{6}$
New zeroes are $3, -\frac{1}{2}$
Sum of new zeroes = $3 + (-\frac{1}{2}) = \frac{5}{2}$
Product of new zeroes = $3 \times (-\frac{1}{2}) = -\frac{3}{2}$
$\therefore$ Required polynomial is $x^2 - \frac{5}{2}x - \frac{3}{2}$ or $2x^2 - 5x - 3$
$\therefore$ Zeroes are $1, -\frac{1}{6}$
New zeroes are $3, -\frac{1}{2}$
Sum of new zeroes = $3 + (-\frac{1}{2}) = \frac{5}{2}$
Product of new zeroes = $3 \times (-\frac{1}{2}) = -\frac{3}{2}$
$\therefore$ Required polynomial is $x^2 - \frac{5}{2}x - \frac{3}{2}$ or $2x^2 - 5x - 3$