8(3x−2y)2−6x+4y−1
(12x−8y+1)(6x−4y−1)
Given 8(3x−2y)2−6x+4y−1
8(3x−2y)2−2(3x−2y)−1 ( taking 2 common in the second term )
Let 3x−2y=a
⇒8a2−2a−1
⇒8a2−4a+2a−1 ( splitting the middle term as -4a and 2a )
⇒4a(2a−1)+1(2a−1)
⇒(4a+1)(2a−1)
Re-substituting a=3x−2y
⇒(4(3x−2y)+1)(2(3x−2y)−1)
⇒(12x−8y+1)(6x−4y−1)