How do you factor $4 - 81 w^{2}$ $4-81w^2$ ?

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Answer 1 · 2016-06-16T10:28:18

$4 - 81 w^{2} = (2 + 9 w) (2 - 9 w)$ $4-81w^2=(2+9w)(2-9w)$

We use the identity $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$

Hence $4 - 81 w^{2}$ $4-81w^2$

= $2^{2} - {(9 w)}^{2}$ $2^2-(9w)^2$

= $(2 + 9 w) (2 - 9 w)$ $(2+9w)(2-9w)$

How do you factor 4−81w24-81w^2?