How do you factor completely $81x^2 – 100$ ?

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Answer 1 · 2016-06-10T15:15:26

$(9x+10)(9x-10)$

To factor $81x^2-100$ we must first recognize that this is the difference (subtraction) of two perfect squares.

$81x^2 = (9x)(9x)$
$100=(10)(10)$

When we factors the difference of two squares the binomials always have one $+$ and one $-$ .

$(9x+10)(9x-10)$

How do you factor completely 81x^2 – 10081x^2 – 100?