How do you factor completely 64x^4-9 ?
1 Answer
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
=(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)
