Find the zeros of the polynomial verify the relationship between the zeroes and their coefficients:
(a) $x^{2} + \frac{1}{6} x - 2$
(b) $4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3}$
(c) $2 x^{2} - 9 - 3 x$

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Solution

$(a) x^{2} + \frac{1}{6} x - 2 = 0$
$6 x^{2} + x - 12 = 0$
$6 x^{2} + 9 x - 8 x - 12 = 0$
$3 x (2 x + 3) - 4 (2 x + 3) = 0$
$x = \frac{4}{3}, x = \frac{- 3}{2}$

$(b) 4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3}$
$4 \sqrt{3} x^{2} + 8 x - 3 x - 2 \sqrt{3} = 0$
$4 x (\sqrt{3} x + 2) - \sqrt{3} (\sqrt{3} x + 2)$
$(4 x - \sqrt{3}) (\sqrt{3} x + 2) = 0$
$x = \frac{\sqrt{3}}{4}, x = \frac{- 2}{\sqrt{3}}$

$(c) 2 x^{2} - 3 x - 9 = 0$
$2 x^{2} - 6 x + 3 x - 9 = 0$
$2 x (x - 3) + 3 (x - 3) = 0$
$(2 x + 3) (x - 3) = 0$
$x = \frac{- 3}{2}, x = 3$

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