Please solve q 14 is of Factorisation
4(2x-3y)^2-8x+12y-3
= 4(2x-3y)^2-4(2x-3y)-3
Substituting (2x-3y) as A,
We get
4A^2-4A-3
= 4A^2-6A+2A-3
= 2A(2A-3)+1(2A-3)
= (2A+1)(2A-3)
Putting the value (2x-3y) in place of A,
We get
{2(2x-3y)+1}{2(2x-3y)-3}
= (4x-6y+1)(4x-6y-3)
Hope this helps.
= 4(2x-3y)^2-4(2x-3y)-3
Substituting (2x-3y) as A,
We get
4A^2-4A-3
= 4A^2-6A+2A-3
= 2A(2A-3)+1(2A-3)
= (2A+1)(2A-3)
Putting the value (2x-3y) in place of A,
We get
{2(2x-3y)+1}{2(2x-3y)-3}
= (4x-6y+1)(4x-6y-3)
Hope this helps.