2y(y^2-3y-2)-3(5y+2y^2-y^3)

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2y(y^2-3y-2)-3(5y+2y^2-y^3)

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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ok so you need to distribute out each term
so 1. 2y(y^2-3y-2) = (2y*y^2) - (2y *3y) - (2y*2) = 2y^3 -6y^2 -4y
id "factor" out each term; but feel free to distribute them out :)

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hmmm, good point
thank u!
well I'll do it this way, then you sow that way =)
2. 3(5y+2y^2-y^3) = (3*5y) + (3*2y^2) - (3*y^3) = 15y + 6y^2 - 3y^3
youre doing fine :)
now subtract them: 2y^3 -6y^2 -4y - (15y + 6y^2 - 3y^3) = 2y^3 -6y^2 -4y -15y -6y^2 +3y^3
and combine like terms (all of the y's, y^2 and y^3): -y^3 -12y^2 -19y
oops one second!
first one is 5y^3 heh, since 2+3=5, not -1 =)
so I get 5y^3 -12y^2 -19y

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