How do you factor completely: $81 x^{4} - 16$ $81x^4 - 16$ ?

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Answer 1 · 2015-07-17T07:32:52

$= (9 x^{2} + 4) (3 x + 2) (3 x - 2)$ $=color(red)((9x^2+4)(3x+2)(3x-2))$

-As per the identity:

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

Here :
$9 x^{2} = a$ $color(blue)(9x^2) = a$
$4 = b$ $color(blue)(4) =b$

Applying the identity to the expression:
$81 x^{4} - 16 = {(9 x^{2})}^{2} - 4^{2}$ $81x^4 - 16 =color(blue)( (9x^2)^2 - 4^2$

$= (9 x^{2} + 4) (9 x^{2} - 4)$ $=color(blue)((9x^2+4))color(green)((9x^2-4)$

$9 x^{2} - 4$ $9x^2-4$ can be factorised further using the same identity

$9 x^{2} - 4 = {(3 x)}^{2} - 2^{2} = (3 x + 2) (3 x - 2)$ $9x^2-4 = (3x)^2 - 2^2 = color(green)((3x+2)(3x-2)$

$81 x^{4} - 16 = (9 x^{2} + 4) (3 x + 2) (3 x - 2)$ $81x^4 - 16 =color(red)((9x^2+4)(3x+2)(3x-2))$

How do you factor completely: 81x^4 - 1681x4−16 81x^4 - 16?