How do you factor completely #64x^4-9# ?

1 Answer
George C.
Aug 2, 2016

#64x^4-9#

#=(8x^2-3)(8x^2+3)#

#=(2sqrt(2)x-sqrt(3))(2sqrt(2)x+sqrt(3))(8x^2+3)#

#=(2sqrt(2)x-sqrt(3))(2sqrt(2)x+sqrt(3))(2sqrt(2)x-sqrt(3)i)(2sqrt(2)x+sqrt(3)i)#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We will use this three times below.

#64x^4-9#

#=(8x)^2-3^2#

#=(8x^2-3)(8x^2+3)#

If we allow irrational coefficients then we can factor this further:

#=((2sqrt(2)x)^2-(sqrt(3))^2)(8x^2+3)#

#=(2sqrt(2)x-sqrt(3))(2sqrt(2)x+sqrt(3))(8x^2+3)#

That is as far as we can go with Real coefficients, since #8x^2+3 > 0# for all Real values of #x#. If we allow Complex coefficients then we can factor this further:

#=(2sqrt(2)x-sqrt(3))(2sqrt(2)x+sqrt(3))((2sqrt(2)x)^2-(sqrt(3)i)^2)#

#=(2sqrt(2)x-sqrt(3))(2sqrt(2)x+sqrt(3))(2sqrt(2)x-sqrt(3)i)(2sqrt(2)x+sqrt(3)i)#

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