How do you factor $9 y^{2} - 1$ $9y^2 - 1$ ?

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How do you factor $9 y^{2} - 1$ $9y^2 - 1$ ?

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Answer 1 · 2018-03-19T11:55:38

$9 y^{2} - 1 = (3 y + 1) (3 y - 1)$ $9y^2-1=(3y+1)(3y-1)$

Explanation:

$9 y^{2} - 1$ $9y^2-1$
$= {(3 y)}^{2} - {(1)}^{2}$ $=(3y)^2-(1)^2$
Using the identity $a^{2} - b^{2} = (a + b) (a - b),$ $a^2-b^2=(a+b)(a-b),$
You can further simplify $({(3 y)}^{2} - {(1)}^{2})$ $((3y)^2-(1)^2)$ as
${(3 y)}^{2} - {(1)}^{2} = (3 y + 1) (3 y - 1)$ $(3y)^2-(1)^2=(3y+1)(3y-1)$

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How do you factor $9 y^{2} - 1$ $9y^2 - 1$ ?

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