Question

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

Answer 1

`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)`