The H.C.F of $12, 14 x^{2} y, 2 y z^{2}$ is

2 z^{2}

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2 x^{2}

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2

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2 y z^{2}

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Solution

The correct option is C $2$
$12, 14 x^{2} y, 2 y z^{2}$
$12 = 2 \times 2 \times 3$
$14 x^{2} y = 2 \times 7 \times x y$
$2 y z^{2} = 2 \times y z^{2}$
So, the common factor is $2$ .
So, the H.C.F of $12, 14 x^{2} y, 2 y z^{2}$ is $2$ .

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