How do you factor completely $6 y^{2} - 54$ $6y^2 - 54$ ?

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How do you factor completely $6 y^{2} - 54$ $6y^2 - 54$ ?

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Answer 1 · 2016-04-22T20:38:53

$6 (y - 3) (y + 3)$ $6(y-3)(y+3)$

Explanation:

$6 y^{2} - 54 = 6 y^{2} - 6 \cdot 9 = 6 (y^{2} - 9) = 6 [{(y)}^{2} - {(3)}^{2}] = 6 (y - 3) (y + 3)$ $6y^2-54=6y^2-6*9=6(y^2-9)=6[(y)^2-(3)^2]=6(y-3)(y+3)$

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How do you factor completely $6 y^{2} - 54$ $6y^2 - 54$ ?

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How do you factor completely 6y2−546y^2 - 54?

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How do you factor completely $6 y^{2} - 54$ $6y^2 - 54$ ?