How do you factor $18 y^{2} - 6$ $18y^2 - 6$ ?

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Answer 1 · 2018-04-20T16:54:23

$\Rightarrow 6 (3 y^{2} - 1)$ $=>6(3y^2-1)$

$\Rightarrow 18 y^{2} - 6$ $=>18y^2-6$

The greatest common factor of $18$ $18$ and $6$ $6$ is $6$ $6$ , so we can factor that out.

$\Rightarrow 6 (3 y^{2} - 1)$ $=>6(3y^2-1)$

This cannot be simplified further. So the final factored form is:

$\Rightarrow 6 (3 y^{2} - 1)$ $=>6(3y^2-1)$

How do you factor 18y^2 - 6 18y2−618y^2 - 6 ?