factorise 10xy(14y^2+43y-21)/5x(7y-3)

10xy(14y² + 43y - 21)/(5x(7y -3))

=2y(14y² + 43y - 21)/(7y -3)  # 5x is common between numerator and denominator
Let us factorise 14y² + 43y - 21
one of them must be 7y -3

2y(7y) +2y(-3) +7(7y) +7(-3)
= (2y + 7)(7y - 3)

Thus 2y(14y² + 43y - 21)/(7y -3) 
=2y( (2y + 7)(7y - 3)) /(7y -3)  
=2y*(2y+7)