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

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Answer 1 · 2015-07-17T07:32:52

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

-As per the identity:

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

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

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

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

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

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

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

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