Find the zeros of $6 x^{2} - 3 - 7 x$

\frac{3}{2}, \frac{1}{3}

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- \frac{3}{2}, - \frac{1}{3}

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- \frac{3}{2}, \frac{1}{3}

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\frac{3}{2}, - \frac{1}{3}

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Solution

The correct option is D: $\frac{3}{2}, - \frac{1}{3}$
Here, $f (x) = 6 x^{2} - 3 - 7 x$
$= 6 x^{2} - 7 x - 3$
$= 6 x^{2} - 9 x + 2 x - 3$
$= 3 x (2 x - 3) + 1 (2 x - 3)$
$= (2 x - 3) (3 x + 1)$ .
Then, $f (x) = 0$
$⟹ 2 x - 3 = 0$ or, $3 x + 1 = 0$ .
Therefore, $2 x = 3$ or, $3 x = - 1$ .
That is, $x = \frac{3}{2}$ , $x = \frac{- 1}{3}$ .
The zeros of the given polynomials are $\frac{3}{2}$ , $\frac{- 1}{3}$ .

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