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Find the quadratic polynomial the sum of whose zeroes is $1$ and their product is $-12$. Hence find the zeroes of the polynomial.
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Sum of zeroes = $1$, Product of zeroes = $-12$ ($\frac{1}{2}$ Mark)
Required polynomial = $(x^2 - x - 12)$ (1 Mark)
$= (x - 4)(x + 3)$ ($\frac{1}{2}$ Mark)
Equating to zero, $x = 4, -3$
$\therefore$ Zeroes are $4$ and $-3$ ($\frac{1}{2}$ Mark)
Required polynomial = $(x^2 - x - 12)$ (1 Mark)
$= (x - 4)(x + 3)$ ($\frac{1}{2}$ Mark)
Equating to zero, $x = 4, -3$
$\therefore$ Zeroes are $4$ and $-3$ ($\frac{1}{2}$ Mark)