Find the zeroes of the following quadratic polynomial: $6 x^{2} + x - 5$

\frac{1}{2}, 1

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6, 5

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\frac{5}{6}, - 1

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2, 1

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Solution

The correct option is C $\frac{5}{6}, - 1$
$6 x^{2} + x - 5 = 0$
$6 x^{2} + 6 x - 5 x - 5 = 0$
$6 x (x + 1) - 5 (x + 1) = 0$
$(6 x - 5) (x + 1) = 0$
So,
$6 x - 5 = 0$
$\Rightarrow x = \frac{5}{6}$

or,
$x + 1 = 0$
$\Rightarrow x = - 1$
Thus, the zeroes of the polynomial $6 x^{2} + x - 5$ are $\frac{5}{6}$ and -1.

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