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