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

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Answer 1 · 2016-06-10T15:15:26

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

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

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

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

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

How do you factor completely 81x2–10081x^2 – 100?