Question

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

Answer 1

ok so you need to distribute out each term

Answer 2

so
1. 2y(y^2-3y-2) = (2y*y^2) - (2y *3y) - (2y*2) = 2y^3 -6y^2 -4y

Answer 3

id "factor" out each term; but feel free to distribute them out :)

Answer 4

hmmm, good point

Answer 5

thank u!

Answer 6

well I'll do it this way, then you sow that way =)

Answer 7

2. 3(5y+2y^2-y^3) = (3*5y) + (3*2y^2) - (3*y^3) = 15y + 6y^2 - 3y^3

Answer 8

youre doing fine :)

Answer 9

now subtract them:
2y^3 -6y^2 -4y - (15y + 6y^2 - 3y^3) =
2y^3 -6y^2 -4y -15y -6y^2 +3y^3

Answer 10

and combine like terms (all of the y's, y^2 and y^3):
-y^3 -12y^2 -19y

Answer 11

oops one second!

Answer 12

first one is 5y^3 heh, since 2+3=5, not -1 =)

Answer 13

so I get 5y^3 -12y^2 -19y