The zeroes of the polynomial $6 x^{2} + 17 x - 88$ are

- 2, \frac{9}{2}

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- \frac{8}{3}, \frac{11}{2}

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2, - \frac{9}{2}

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- \frac{11}{2}, \frac{8}{3}

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Solution

The correct option is B $- \frac{11}{2}, \frac{8}{3}$
$6 x^{2} + 17 x - 88$
Using common factor form,
$= 6 x^{2} + 33 x - 16 x - 88$
$= 3 x (2 x + 11) - 8 (2 x + 11)$
$= (3 x - 8) (2 x + 11)$
Since, the factor are $(3 x - 8) & (2 x + 11)$ , to find the zeroes, we will equate them to zero
$\Rightarrow (3 x - 8) = 0$ $&$ $(2 x + 11) = 0$
$\Rightarrow x = \frac{8}{3}$ $&$ $x = - \frac{11}{2} .$
Hence, the answer is $x = \frac{8}{3}$ $&$ $x = - \frac{11}{2} .$

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