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)#

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)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Julia Adams

To factor completely ( 64x^4 - 9 ), you can use the difference of squares formula, which states that ( a^2 - b^2 = (a + b)(a - b) ). Applying this formula to your expression, we get: ( 64x^4 - 9 = (8x^2 + 3)(8x^2 - 3) ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.