Factorise: (2x - 5)^2 - 4?

(2x - 5)^2 - 4
= (2x - 5)(2x - 5) - 4
= 4x^2 - 10x - 10x + 25 - 4
= 4x^2 - 20x + 21
= (2x - 7)(2x - 3)

[ (2x - 5) - 2 ] [ (2x - 5) + 2 ]
(2x - 7) (2x - 3)

(2x - 5)^2 - 4
=(2x-5+2)(2x-5-2)
=(2x-3)(2x-7)

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

(2x - 5)^2 - 4 = (2x - 5)^2 - 2^2
= (2x-5 - 2)(2x-5 + 2)
= (2x - 7)(2x - 3)

The format of factoring a difference of squares is the following:

a^2 - b^2 factors as (a - b)(a + b).

With that said, do the exact same thing here, since (2x - 5)^2 is clearly square, and 4 is square (in fact it is 2 squared). Showing you in steps though,

(2x - 5)^2 - 4
(2x - 5)^2 - 2^2

( (2x - 5) - 2 ) ( (2x - 5) + 2)

And just simplify the term inside the brackets.

(2x - 7)(2x - 3)