How do you factor completely $4 y^{2} - 36$ $4y^2-36$ ?

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Answer 1 · 2016-05-15T12:49:21

$4 (y - 3) (y + 3)$ $4(y - 3) (y+3)$

$4 y^{2} - 36$ $4y^2 - 36$

= $4 (y^{2} - 9)$ $4(y^2 - 9)$

= $4 (y - 3) (y + 3)$ $4 (y-3)(y+3)$

How do you factor completely 4y^2-364y2−364y^2-36?