## Algebra 1

$6y(10y^{2}-7)(y-5)$
Given : $6y(10y^{3}-50y^{2}-7y+35)$ This becomes : $=6y(10y^{2}(y-5)-7(y-5))$ $=6y((10y^{2}-7)(y-5))$ $=6y(10y^{2}-7)(y-5)$