How do you factor $81 x^{2} - 144$ $81x ^ { 2} - 144$ ?

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Answer 1 · 2017-01-27T16:40:46

The answer is $= 9 (3 x + 4) (3 x - 4)$ $=9(3x+4)(3x-4)$

We use

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

Therefore,

$81 x^{2} - 144 = 9 (9 x^{2} - 16)$ $81x^2-144 =9(9x^2-16)$

$= 9 (3 x + 4) (3 x - 4)$ $=9(3x+4)(3x-4)$

How do you factor 81x ^ { 2} - 14481x2−14481x ^ { 2} - 144?