I need help expanding this equation

(2x - 1)^4

(2x-1)(2x-1)(2x-1)(2x-1)
= (4x^2 - 4x + 1) (2x-1) (2x-1)
= (8x^3 - 4x^2 - 8x^2 + 4x + 2x - 1) ( 2x-1)
= (8x^3 -12x^2 + 6x -1) (2x-1)
= 16x^4 - 8x^3 - 24x^3 + 12x^2 + 12x^2 - 6x - 2x + 1
= 16x^4 - 32x^3 + 24x^2 - 8x + 1

(2x-1)^4 = (2x-1)(2x-1)(2x-1)(2x-1)

Then you can multiply them out.

Two binomials at a time.

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

(4x^2 - 4x + 1) (4x^2 - 4x + 1)

Then multiply these trinomials.

Just multiply 1 term at a time.

(2x-1)^4 = (2x-1)*(2x-1)*(2x-1)*(2x-1)

= (4x^2-4x+1)*(2x-1)*(2x-1) or you could write it as (4x^2-4x+1)* (4x^2-4x+1)

= 16x^4 -16x^3 +4x^2 - 16x^3 +16x^2 -4x +4x^2 -4x +1 (now combine like terms or terms with the same exponent)

= 16x^4 -32x^3 +24x^2 -8x + 1

You can plug in (2x-1)^3, since ( )^3 is given for binomial form.
Then, multiply the result by (2x-1).

First make it
(2x-1)^4 = (2x-1)(2x-1)(2x-1)(2x-1)

Then you can multiply them.