Factorization 急急急急急

Answer: ( 2 + a – 2b )( 2 – a + 2b )

Solution:
4 – a^2 – 4b^2 + 4ab
= 4 – ( a^2 + 4b^2 – 4ab )
= 4 – ( a^2 – 4ab + 4b^2 )
= 4 – [( a )^2 – 2( a )( 2b ) + ( 2b )^2 ]
= 4 – ( a – 2b )^2
= ( 2 )^2 – ( a – 2b )^2
= [( 2 ) + ( a – 2b )] [( 2 ) – ( a – 2b )]
= ( 2 + a – 2b )( 2 – a + 2b )

4-a²-4b²+4ab.
=4-[a²-4ab+(2b)²]
=2²-(a-2b)²
=(2+a-2b)(2-a+2b)

since there is a^2 & b^2 and also integer in the formula, let's assume the formula can be factorize as

(xa + yb + z) (x'a + y'b +z')
= xx' a^2 + yy' b^2 + (xy' + x'y) ab + (xz' + x'z) a + (yz' + y'z) b + zz'

Therefore.
xx' = -1
yy' = -4
xy'+x'y = 4
xz'+x'z = 0
yz'+y'z = 0
zz' = 4

Since xx' = -1, so either x & x' is 1 or -1. Let's x=1, x'=-1
xz' + x'z = 0, so z = z' & because zz' = 4, so z = z' = 2 or -2
Since yz' + y'z = 0, so y = -y' & because yy' = -4 so y & y' is 2 or -2
Since xy' + x'y = 4, so y' is 2 & y is -2

So after factorize, the formula become

(a-2b+2) * (-a+2b+2) or
(a-2b-2) * (-a+2b-2)