## anonymous 5 years ago 3y^3 (2y-5)(y+5)

1. anonymous

What?

2. anonymous

$3y ^{3}\times(2y-5)(y+5) = 3y ^{3}\times(2y ^{2}+10y-5y-25)=6y ^{5}+15y ^{4}-75y ^{2}$

3. anonymous

ops the last term should be to the third not second: $-75y ^{3}$

4. anonymous

yes that is one of the options & thank you! I didn't realize that they could be separated like this 3y3×(2y−5)(y+5) & 3y3×(2y2+10y−5y−25)