factorise 25-(x-2)^2?

simple
5^2 - (x-2)^2 ---------> a^2 - b^2 = (a - b)(a + b)
= (5 - (x-2))(5 + (x -2))
= (7 - x) (3 + x)

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

That's a difference of squares (a² - b²) which factorizes as:
(a - b)(a + b):

In your case a = 5, b = x-2:
(5 - (x-2))(5 + (x-2))

Simplifying:
(5 - x + 2)(5 + x - 2)
(7 - x)(3 + x)

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

25 - (x-2)^2
= 25 - (x^2 - 4x + 4)
dy/dx = -2x + 4

think that's right...

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