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))#