How do you factor $81 x^{2} - 1$ $81x^2-1$ ?

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Answer 1 · 2015-07-17T07:34:23

$= (9 x + 1) (9 x - 1)$ $=color(blue)((9x+1)(9x-1)$

-As per the identity:

$(a^{2} - b^{2}) = (a + b) (a - b)$ $color(blue)((a^2-b^2) = (a+b)(a-b)$

Here :
$9 x = a$ $color(blue)(9x) = a$
$1 = b$ $color(blue)(1) =b$

Applying the identity to the expression:
$81 x^{2} - 1 = {(9 x)}^{2} - 1^{2}$ $81x^2 - 1 =color(blue)( (9x)^2 -1^2$

$= (9 x + 1) (9 x - 1)$ $=color(blue)((9x+1)(9x-1)$

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