Answer :
Answer will help you.
2(3y − 1)(y − 3)
=(2(3y−1))(y+−3)
=(2(3y−1))(y)+(2(3y−1))(−3)
=6y^2−2y−18y+6
Answer is =6y^2−20y+6
2(3y − 1)(y − 3)
=(2(3y−1))(y+−3)
=(2(3y−1))(y)+(2(3y−1))(−3)
=6y^2−2y−18y+6
Answer is =6y^2−20y+6