4(x^2-4x+3)
You can factor this further by seeing which 2 numbers subtract to 4 (your middle term) and add to 3 (your last term). Post if you want it checked/need help.
y=4x^2-16x+12
3 answers
4[x^2-3x-x+3]
=4[x[x-3)-1(x-3)]
=4[x-1][x-3]
=4[x[x-3)-1(x-3)]
=4[x-1][x-3]
4[x-1][x-3]